Vedic Mathematcis-1

Tips To Remember Some Values

The values of reciprocal percentages (RP) for 6 is exactly half that for 3 (half of 33.33 = 16.66) The RP for 8 is exactly half of 4 (half of 25 = 12.5)

Seven is easy to remember just 7 into 2 (14), followed by 14 into 2 (28) which makes it 14.28

9 is one-third of 3 (33.33 divided by 3 = 11.11)

Please start with the next ten only after becoming absolutely comfortable with the first ten

By now you would have figured out that the difficult ones are the prime numbers. We have already dealt with 7. Now we need to work out 11, 13, 17, 19, 23 and 29.

9 and 11 are interrelated as 1/9 is 11.11 and 1/11 is 9.09

13 is considered unlucky. The way you remember it is through the year 1977,which proved unlucky for Indira Gandhi and Coca Cola.

All even numbers can be worked out by dividing the RP for the number that was their half or quarter by two or four respectively. For example, 12 is half of 6 (half of 16.66 = 8.33)

Workout the rest of the primes and your own unique way to remember them. We cannot emphasize the importance of having the percentages of the 1st 30 reciprocals on your fingertips.

Miscellaneous Topics

Coin Problems

The key to these problems is to keep the quantity of coins distinct from the value of the coins. An example will illustrate.

Example : Laura has 20 coins consisting of quarters and dimes. If she has a total of $3.05, how many dimes does she have?

(A) 3 (B) 7 (C) 10 (D) 13 (E) 16

Let D stand for the number of dimes, and let Q stand for the number of quarters. Since the total number of coins in 20, we get D + Q = 20, or Q = 20 - D. Now, each dime is worth 10 cents, so the value of the dimes is 10D. Similarly, the value of the quarters is 25Q = 25(20 - D). Summarizing this information in a table yields

Notice that the total value entry in the table was converted from $3.05 to 305 cents. Adding up the value of the dimes and the quarters yields the following equation:

10D + 25(20 - D) = 305
10D + 500 - 25D = 305
-15D = -195
D = 13

Hence, there are 13 dimes, and the answer is (D).

Maximum Value of an expression

(A^x ) (B^y) (C^z) will be maximum when A/x = B/y = C/z

Ex (a+x)^3 (a-y)^4 will have the maximum value when

(a+x)/3 = (a-y)/4

i.e 4a+4x=3a-3y i.e a=3y+4x

Age Problems

Typically, in these problems, we start by letting x be a person's current age and then the person's age a years ago will be x - a and the person's age a years in future will be x + a. An example will illustrate.

Example : John is 20 years older than Steve. In 10 years, Steve's age will be half that of John's. What is Steve's age?

(A) 2 (B) 8 (C) 10 (D) 20 (E) 25

Steve's age is the most unknown quantity. So we let x = Steve's age and then x + 20 is John's age. Ten years from now, Steve and John's ages will be x + 10 and x + 30, respectively. Summarizing this information in a table yields

Since "in 10 years, Steve's age will be half that of John's," we get

(x + 30)/2 = x + 10
x + 30 = 2(x + 10)
x + 30 = 2x + 20
x = 10

Hence, Steve is 10 years old, and the answer is (C).

Prime Numbers

How many divisors does 1200193 have and what is the sum of all those divisors?

Ans: (From the post http://www.cat4mba.com/node/3533#comment-1255 - user rajorshi)

The big pain being to fing the prime factors of 1200193 which i wont take, who knows the number could be prime and iwould have wasted half  a day.

let me give the formula generally :

if p = a^m  * bⁿ  * c^o  * ..... where a,b,c are prime.

thrn the number of divisors = (m+1)(n+1)(o+1).....

the sum of divisors = [(a^m+1  - 1)(bⁿ⁺¹  - 1)(c^o+1  - 1)......] / [(a-1)(b-1)(c-1).....]

Number System

A Number is an abstract entity that represents a count or measurement. All numbers fall in 2 categories Real Number, Complex Number A Real Number can either be a Rational number or an irrational number. Rational number can either be Natural numbers or Negative integers or Fractions Natural Numbers
The natural numbers start off as follows: 1, 2, 3, 4, and 5 ... The "..." means that the list goes on forever. We give this set the name N. If a number is in N, then its successor is also in N. Thus, there is no greatest number, because we can always add one to get a larger one. N is an infinite set . Since it is infinite, N can never be exhausted by removing its members one at a time. Whole Numbers
If we add zero to our above list then we have the set of whole numbers. i.e Whole numbers 0,1,2,3... Negative numbers
Negative numbers are numbers which are less than zero. They are used to indicate a number that is opposite to the corresponding positive number (the absolute value), but equal in magnitude. Exampe: -1, -2, -3, . . . Remember -(n + 1)is always smaller than -n where n is a positive number. Integer
Integers are the whole numbers, negative whole numbers, and zero. One of the numbers ..., -2, -1, 0, 1, 2, . . . But numbers like 1/2, 4.00032, 2.5, Pi, and -9.90 are not integers. Note that zero is neither positive nor negative. It may help you to think of numbers as occurring along a line that stretches infinitely in both directions. Numbers to the left of the 0 point are negative, numbers to the right are positive. Along this line are a series of dots that correspond to whole numbers (integers). The spaces between the whole numbers are occupied by the numbers that are not whole (they contain fractions, and are called real numbers ex.1/2, 4.00032, 2.5, Pi, and -9.90). Even and Odd
The terms even and odd only apply to integers. A number is said to be an even number if it is divisible by 2 or else it is an odd number. Even numbers are: 2, 4, 6, 8, 10. . . . .40, 42, 44,. . . 312, 314, .... 1008,1010, . . . .686860.... Odd numbers are: . . 5, 7, 9. . . . .41, 43, 45,. . . 311, 313, .... 1007,1009, . . . .686861.... 2.5 is neither even nor odd. Zero, on the other hand, is even since it is 2 times some integer: it's 2 times 0. To check whether a number is odd, see whether it's one more than some even number: 7 is odd since it's one more than 6, which is even. Another way to say this is that zero is even since it can be written in the form 2*n, where n is an integer.Odd numbers can be written in the form 2*n + 1. Again, this lets us talk about whether negative numbers are even and odd: -9 is odd since it's one more than -10, which is even. Every positive integer can be factored into the product of prime numbers, and there's only one way to do it for every number . For instance, 280 = 2x2x2x5x7, and there's only one way to factor 280 into prime numbers Rational Number
A rational number is a number that can be expressed as a fraction p/q where p and q are integers and q ≠ 0. i.e Rational numbers are simply defined as ratios of integers. 1/2 is a rational number. 2/3 is also a rational number. Note that all Of the integers are rational numbers, because you can think of them as the ratio of themselves to 1, as in 2 = 2/1 which is certainly the ratio of two integers, and so 2 is a rational number. The decimal form of a rational number is either a terminating or repeating decimal. Representation of rational numbers in decimal form Any positive rational number p/q, after actual division, if necessary can be expressed as, p / q = m + r/q where m is non-negative integer and 0 ≤ r < q For example , 31/5 = 6 + 1/5 = 6.2 There are few fractions for which the right most digit(or set of right most digit)recurrs endlessly. For example 1/3 =0.33333. . . . and 5/11 = 0.45454 . . . . Note that the dots ........ represent endless recurrence of digits. The above examples are decimal numbers of the "non-terminating type". In case of "non-terminating type" we have decimal fractions having an infinite number of digits. Some decimal fractions from this group have digits repeating infinitely. They are called "repeating or recurring " decimals. In "endless recurring or infinite repeating" decimal fractions we can see that when p is actually divided by q the possible remainders are 1, 2, 3, ..... , q -1. So one of them has to repeat itself in q steps. Thereafter the earlier numeral or group of numerals must repeat itself. All the rational numbers thus can be represented as a finite decimal (terminating type) or as a recurring decimal. Irrational Numbers
In mathematics, an irrational number is any real number that is not a rational number i.e., one that cannot be written as a ratio of two integers, i.e., it is not of the form a/b where a and b are integers and b is not zero. It can readily be shown that the irrational numbers are precisely those numbers whose expansion in any given base (decimal, binary, etc) never ends and never enters a periodic pattern. The square root of 2 is a classic example of an irrational number: you cannot write it as the ratio of ANY two integers. Prime number
A prime number is a whole number that is not the product of two smaller numbers. Note that the definition of a prime number doesn't allow 1 to be a prime number : 1 only has one factor, namely 1. Prime numbers have exactly two factors, not "at most two" or anything like that. When a number has more than two factors it is called a composite number. Here are the first few prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, etc. PRIME FACTORS Suppose n is a natural number. Then there exists a unique sequence of prime numbers p₁, p₂, p₃, . . . , p_m, such that both of the following statements are true: p₁ ≤ p₂ ≤ p₃ ≤ . . . ≤ p_m p₁ x p₂ x p₃ . . . x p_m The numbers p₁, p₂, p₃, . . . , p_m are called the prime factors of the natural number. Every natural number n has one, but only one, set of prime factors. This is an important principle known as the Fundamental Theorem of Arithmetic.

Number of Prime Factors
A number N of the form
a^m x bⁿ x c^p
where a, b, c are all prime factors of number N

has (m + 1)(n + 1)(p + 1)no. of prime factors

CAT TRICKS

Divisibility Rules

Dividing by 3

Add up the digits: if the sum is divisible by three, then the number is as well.

Examples: 111111: the digits add to 6 so the whole number is divisible by three. 87687687. The digits add up to 57, and 5 plus seven is 12, so the original number is divisible by three.

Dividing by 4

Look at the last two digits. If the number formed by its last two digits is divisible by 4, the original number is as well. Examples: 100 is divisible by 4. 1732782989264864826421834612 is divisible by four also, because 12 is divisible by four.

Dividing by 5

If the last digit is a five or a zero, then the number is divisible by 5.

Dividing by 6

Check 3 and 2. If the number is divisible by both 3 and 2, it is divisible by 6 as well.

Dividing by 7

To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number.

Example: If you had 203, you would double the last digit to get six, and subtract that from 20 to get 14. If you get an answer divisible by 7 (including zero), then the original number is divisible by seven. If you don't know the new number's divisibility, you can apply the rule again.

TEST Take the number and multiply each digit beginning on the right hand side (ones) by 1, 3, 2, 6, 4, 5. Repeat this sequence as necessary .Add the products. If the sum is divisible by 7 - so is your number.Example: Is 2016 divisible by 7? 6(1) +1(3) + 0(2) + 2(6) = 21 . 21 is divisible by 7 and we can now say that 2016 is also divisible by 7.

Dividing by 8

Check the last three digits. Since 1000 is divisible by 8, if the last three digits of a number are divisible by 8, then so is the whole number.

Example: 33333888 is divisible by 8; 33333886 isn't.

Dividing by 9

Add the digits. If that sum is divisible by nine, then the original number is as well.

Example:  12348 is divisible by 9; as the sum is 18

Dividing by 10

If the number ends in 0, it is divisible by 10.

Example: 20 , 345ABCV80

Dividing by 11

Find the sum of alternate numbers and if the differnce between those two numbers is zero or multiple of 11 then the number is divisible by 11.

Let's look at 352, which is divisible by 11;  the answer is 3+2 is 5 and 5 - 5 = 0

another way  to say this is that 35 -2 is 33.

Dividing by 12

Check for divisibility by 3 and 4.

Example:  936, it is divisible by both 3 and 4 and thus by 12.

Dividing by 13

Delete the last digit from the given number. Then subtract nine times the deleted digit from the remaining number. If what is left is divisible by 13, then so is the original number.

Example:  104
10 - 36 = -26, which is divisible by 13 so 104 is divisible by 13

Dividing by 17

How can you know if a number with three or more digits is divisible by the number fourteen?

Check if the last digit of the original number is odd or even. If the number is odd, then the number is not divisible by fourteen. If the number is even, then apply the Dividing by 17 The result of subtracting five times the last digit from the number with the last digit removed is divisible by 17. Example: 187: "18" - ("7" x 5) = -17 is divisible by 17

Dividing by 19

The result of adding twice the last digit to the number with the last digit removed is divisible by 19.

Example: 437: "43" + ("7"x2) = 57 is divisible by 19.

Dividing by 23

The result of adding seven times the last digit to the number with the last digit removed is divisible by 23.

Example: 598: "59"+("8" x 7)= 115 is divisible by 23. TIP: If a number is divisible by two different prime numbers, then it is divisible by the products of those two numbers. Since 36, is divisible by both 2 and 3, it is also divisible by 6.

LCM & HCF

Prime factorisation

Percentages

The term percent comes from Latin and means "for every hundred". So when you hear a statistic such as "10% of all people are left-handed" that means, for every hundred people, 10 are left-handed.
Percent of change (percent of increase or percent of decrease) can be calculated using the following formula:

Example: Richard's health insurance premium for last year was $1440. If he paid $1512 this year, what is the percent of increase on his health insurance premium?
Answer:($1512-$1440)/$1440 — 100% = 72/1440 — 100% = 5% Hence, the percent of increase on the health insurance premium is 5%.

Example: A T-shirt is sold at 20% off the original price of $32. What is the sales price?
Answer: Let the sales price be x dollars. ($32-x)/$32 — 100% = 20% ($32-x)/$32 = 0.2 $32 - x = $6.4 x = $25.6 Hence, the sales price of the T-shirt is $25.6.
 

The percent of a number can be found based on the type of the question asked.

Type 1: x is y percent of n
Translation 1: x = y% — n
Example: 24 is 15% of what number?
Answer: 24 = 15% — n; n = 24 · 15% = 160

Type 2: x percent of y is n
Translation 2: x% — y = n
Example: 11% of 50 is what number?
Answer: 11% — 50 = n; n = 5.5

  CATrick 
What is 320% of 25?
Many of the students will start doing like this:(320/100)*25 , little realizing that 320% of 25 is same as 25% of 320. so the better and faster method is to deal with fractions and this question is best solved like this: (25/100)*320 i.e 1/4th of 320 and hence answer is 80.

A common error when using percentages is to imagine that a percentage increase is cancelled out when followed by the same percentage decrease.
A 50% increase from 100 is 100 + 50, or 150. A 50% reduction from 150 is 150 - 75, or 75. The end result is smaller than the 100 we started out with. This phenomenon is due to the change in the "initial" value after the first calculation. In this example, the first initial value is 100, but the second is 150.In general, the net effect is:
(1 + x) (1 - x) = 1 - x²,
that is a net decrease proportional to the square of the percentage change

Example: If a quantity increases by 20% and then decreases by 10%, then what is the net increase/decrease?
Answer: If we take the quantity to be x, then the answer will be found as: x(1.2)(0.9) = 1.08x and hence 8% increase. This calculation is more tedious than when quantity chosen is 100. When quantity is 100, after the 20% increase, the quantity becomes 120 and a decrease of 10% on this will be a decrease of 12 units. Therefore, the quantity becomes 108 and hence 8% increase.

Example: If Rama's income is 20% more than Shyama's income, by what percentage is Shyama's income less than Rama's?
Answer: Instead of using x or y, take Shyama's income as Rs 100. Then, Ram's income = Rs 120.
Therefore, Shyama's income is Rs 20 less or 1/6th less or 16.66% less than Rama's income.
Similarly, if Ram's income were less by 20%, then Rama's income would be Rs 80 and hence Shyam's income would be Rs 20 more or 25% more than Rama's income.

Note that the percentage is changed because of base (though the quantity of increase or decrease of Shyam's income is same).

Profit & Loss

DEFINITIONS: SALES PRICE (SP): The price at which goods/services are sold COST PRICE (SP): The expenses occurred in making a product (or providing a service) and it includes the price of the raw material. REVENUE: is compensation received for your product or services. COST OF GOODS SOLD: are the variable expenses related to the sale of your product. OPERATING EXPENSES: are fixed expenses, such as rent, and utilities. OPERATING INCOME: is profit after operating income. EARNINGS BEFORE TAXES: is income including other income and expenses, but before taxes. INCOME TAXES: are federal, state, and local taxes. NET EARNINGS: is the profit earned by the business, and it includes all expenses, including taxes. If the shop sells a thing for more than they paid for it, then the difference is the 'profit':
profit = selling price - cost price If the shop sells a thing for less than they paid for it, then the difference is the 'loss':
loss = cost price - selling price Both profit and loss can be expressed either in dollars, OR as a percentage of the cost price.
Profit Percentage = (Profit / Cost Price) * 100 Occasionally profit is also mentioned in terms of sales price i.e
Profit on selling price= (Profit / Sales Price) * 100 Several problem types based on these formulas are discussed in the examples below ========================================================================= Q1. If a merchant offers a discount of 40% on the marked price of his goods and thus ends up selling at cost price, what was the % mark up? ANS: If the merchant offers a discount of 40% on the marked price, then the goods are sold at 60% of the marked price. The question further states that when the discount offered is 40%, the merchant sells at cost price. Therefore, selling @ 40% discount = 60% of marked price (M) = cost price (C) i.e., a mark up 66.66% Q2. A shopkeeper bought an almirah from a wholesale dealer for Rs 4500 and sold it for Rs 6000. Find his profit or loss percent. Answer: Here C.P. of the almirah = Rs 4500 S.P. of the almirah = Rs 6000 Since S.P. > C.P., there is a profit Profit = S.P. – C.P. = Rs 6000 – Rs 4500 = Rs 1500 Profit % = 1500/4500 = 33.33% . Q3. A dealer marks his goods 10% above the cost price and allow 10% discount. What does he gain or loose percent? Q4. A shopkeeper buys scientific calculators in bulk for $15 each. He sells them for $40 each. Q5. A trader professes to sell his goods at a loss of 8% but weights 900 grams in place of a kg weight. Find his real loss or gain per cent ? Answer: The trader professes to sell his goods at a loss of 8%. Therefore, Selling Price = (100 - 8)% of Cost Price or SP = 0.92CP But, when he uses weights that measure only 900 grams while he claims to measure 1 kg. Hence, CP of 900gms = 0.90 * Original CP So, he is selling goods worth 0.90CP at 0.92CP Therefore, he makes a profit of 0.02 CP on his cost of 0.9 CP profit %= 2.22% Q6. Calculate the profit on each calculator in dollars, and as a percentage of the cost price. Answer: Given cost price = $15, selling price = $40 profit = selling price - cost price = $40 - $15 = $25 Expressing the profit as a percentage of the cost price: $profit profit% = ------------------------ × 100% $cost price $25 = -------- × 100% = 166.7% $15 Q7. : A retailer buys a cooler for Rs 3800 and overhead expenses on it are Rs 50. If he sells the cooler for Rs 4400, determine his profit percent. Answer : Here, C.P. of the cooler = Rs (3800 + 50) = Rs 3850 S.P. of the cooler = Rs 4400 Since S.P. > C.P., there is a profit Profit = Rs 4400 – Rs 3850 = Rs 550 Profit % = 14. 27%. Q8. A trader makes a profit equal to the selling price of 75 articles when he sold 100 of the articles. What % profit did he make in the transaction? Answer: Let S be the selling price of 1 article. Therefore, the selling price of 100 articles = 100 S. --(1) The profit earned by selling these 100 articles = selling price of 75 articles = 75 S -- (2) We know that Selling Price (S.P.) = Cost Price (C.P) + Profit -- (3) Selling price of 100 articles = 100 S and Profit = 75 S from (1) and (2). Substituting this in eqn (3), we get 100 S = C.P + 75 S. Hence, C.P = 100 S - 75 S = 25 S. Profit % = 300% Q9.A purchased a car for Rs. 1,35,000/- and spent Rs. 25,000/- for repairs. At what cost the car was sold if he earned 10% profit on it? Q10. A florist bought 360 roses at $10 per dozen. If he sold them at $1.10 each, what is his profit as a percentage of selling price? Q11. A garrison has enough food to last for 24 days. How much longer will the food last if each individual ration have to be reduced if the food is to last 40 days? Q12. A trader buys goods at a 19% discount on the label price. If he wants to make a profit of 20% after allowing a discount of 10%, by what % should his marked price be greater than the original label price? Answer:Let the label price be = Rs.100. The trader buys at a discount of 19%. Hence, his cost = 100 - 19 = 81. He wants to make a profit of 20%. Hence his selling price = 1.2 (81) = 97.2 However, he wants to get this Rs.97.2 after providing for a discount of 10%. i.e. he will be selling at 90% of his marked price. Hence, his marked price M = = 108 which is 8% more than the original label price. Q13. The cost price of 40 articles is equal to the selling price of 35 articles. What is the profit /loss percent? Answer : Let the cost price of one article be $1, then the cost price of 40 articles is $40 The question states that the selling price of 35 articles = cost price of 40 articles = $40 Therefore, now we know the selling price for 35 articles Now, let us find the cost price for 35 articles. We have assumed the cost of 1 article = $1. Therefore, cost price of 35 articles = $35 Therefore, the profit/loss made on 35 articles = S.P of 35 articles - C.P of 35 articles i.e., $40 - $35= $5, or a profit of $5 Hence, % Profit = = = 14.28% profit Q14. A machine is sold at a profit of 10%. Had it been sold for $40 less, a loss of 10% would have been incurred. What is the Cost Price of the machine? Solve on your own or if face any problem then discuss it in our forum http://www.cat4mba.com/forum/ Q15. A trader makes a profit equal to the selling price of 75 articles when he sold 100 of the articles. What % profit did he make in the transaction? Answer : Let S be the selling price of 1 article. Therefore, the selling price of 100 articles = 100 S. -- (1) The profit earned by selling these 100 articles = selling price of 75 articles = 75 S -- (2) We know that Selling Price (S.P.) = Cost Price (C.P) + Profit -- (3) Selling price of 100 articles = 100 S and Profit = 75 S from (1) and (2). Substituting this in eqn (3), we get 100 S = C.P + 75 S. Hence, C.P = 100 S - 75 S = 25 S. Profit % = Q16. A merchant buys two articles for Rs.600. He sells one of them at a profit of 22% and the other at a loss of 8% and makes no profit or loss in the end. What is the selling price of the article that he sold at a loss? Answer: 404.70 Q17. Rajiv sold an article for Rs.56 which cost him Rs.x. If he had gained x% on his outlay, what was his cost? Answer:Rs. 40 Q18. If apples are bought at the rate of 30 for a rupee. How many apples must be sold for a rupee so as to gain 20%? Answer:25 Q19. A mechanic purchased two motor bikes for $7,500 and $6,000 respectively. He sold the first one for a profit of 15% and the second bike for a loss of 20%. How much did he lose or gain on both the transactions together? $60 loss $75 loss $75 profit $80 profit Q20. A school bookshop sells an outdated biology text book for $49.35, making a 6% loss.What was the cost price of the book, and what is the cash value of the loss? Q21. Two merchants sell, each an article for Rs.1000. If Merchant A computes his profit on cost price, while Merchant B computes his profit on selling price, they end up making profits of 25% respectively. By how much is the profit made by Merchant B greater than that of Merchant A? Answer: Rs. 50 Q22. A fruit vendor purchased four branches of bananas consisting of 540 fruits in all for $168.75. He sold 15 dozens at 2 bananas a dollar, another 15 dozens at 3 bananas a dollar and the remaining at 5 bananas a dollar. Find his loss or gain percentage. 10 2/9% loss 13 1/4% profit 10 2/9% profit 13 1/4% loss Q23. By selling a scooter to a customer for Rs 22400 an auto-dealer makes a profit of 12%. Find the cost price of the scooter. Q24.A merchant marks his goods up by 75% above his cost price. What is the maximum % discount that he can offer so that he ends up selling at no profit or loss? Answer: 42.85% Q25. By selling a cycle for Rs 2024, a cycle merchant loses 12%. If he wishes to make a gain of 12%, what will be the selling price of the cycle ? Q26. If the cost price of 15 articles is the same as the selling price of 12 articles, find the gain or loss percent in the transaction. Q27.A trader purchased 5000 coconuts, at $110 per 100 coconuts. 10% of the coconuts were damaged during transportation. He spent $575 for the transportation. If he gained 40% profit, at what price did he sell them? $3.15 each $1.89 each $1.50 each $2.00 each Q28. A merchant marks his goods up by 60% and then offers a discount on the marked price. If the final selling price after the discount results in the merchant making no profit or loss, what was the percentage discount offered by the merchant? Q29. By selling 45 oranges for Rs 160, a women loses 20%. How many oranges should she sell for Rs 112 to gain 20% on the whole ? Q30. A sells an article costing Rs 1000 to B and earns a profit of 6%. B, in turn sells it to C at a loss of 5%. At what price did C purchase the article ? Q31. By selling 90 ball pens for Rs 160, a person loses 20%. How many ball pens should he sell for Rs 96, so as to have a gain of 20% ? Q32. If the selling price of 20 articles is equal to the cost price of 23 articles, find the loss or gain percent. Q33.A watch was sold at a profit of 12%. Had it been sold for Rs 33 more, the profit would have been 14%. Find the cost price of the watch.

SI and CI

Simple Interest Let the annual rate of interest be i (as a fraction, that is 100i percent), the amount of the principal be P, the number of years be n, and the amount after n years be A. Then A = P(1+ni). If you want to know what principal to deposit in order to have an amount A after n years at interest rate i, that principal is called the present value, and is given by P = A/(1+ni). To find the interest rate i, use i = ([A/P]-1)/n. To determine how many compounding periods are needed to reach a given amount, n = ([A/P]-1)/i. Example: Suppose you deposit $6000 in a bank and receive 4% per year simple interest for 7 years. Then the parameters will be principal P = $6000, interest rate per period i = 0.04, and number of periods n = 7. The amount of interest you will have received by the end of 7 years will be Pni = ($6000)(7)(0.04) = $1680, so you will have A = $7680 Compound Interest Let the annual rate of interest be i (as a fraction, that is 100i percent), the amount of the principal be P, the number of years be n, the number of times per year that the interest is compounded be q, and the amount after n years be A. Then A = P(1+[i/q])nq. Then the present value is given by P = A(1+[i/q])-nq. To find the interest rate i, use i = q([A/P]1/nq - 1). To determine how many years are needed to reach a given amount, n = log(A/P)/(q log[1+(i/q)]). ============================================================================== Interest may be compounded quarterly, monthly, weekly, daily, or even more frequently. As the frequency of compounding increases, the amount A increases, but ever more slowly -- in fact it approaches a limit with continuous compounding. The formulas for this situation are found by taking the limit of the formulas above as q increases without bound. They take the form A = Pein, P = Ae-in, i = log(A/P)/(n log[e]), n = log(A/P)/(i log[e]). Here e = 2.718281828459... is the base of natural logarithms. Example: Suppose you deposit $6000 in a bank and receive 4% per year compound interest for 7 years, compounded monthly. The parameters will be principal P = $6000, annual interest rate i = 0.04, number of years n = 7, and number of periods per year q = 12. Over these 7 years the principal will grow over to the amount A, where A = P(1+[i/q])nq, = ($6000)(1+[0.04/12])(7)(12), = ($6000)(1.003333333...)84, = $7935.08. Annual Percentage Rate (APR) The annual percentage rate (APR) is the rate r of simple interest for 1 year, which will be equal to the actual amount of compound interest for that year at rate i compounded q times per year. It is given by the formula r = (1+[i/q])q-1. To find the interest rate i given the APR r, use i = q[(1+r)1/q-1]. The APR is mainly used to compare loans with different interest rates and payment intervals. The lower the APR, the lower the cost of the loan to the borrower. Example: Suppose your credit card charges 18% interest per year, but you have to pay the interest due monthly. What is the annual percentage rate? Here the parameters are rate i = 0.18 and the number of compounding periods q = 12. Then the annual percentage rate (APR) r is given by r = (1+[i/q])q-1, = (1+[0.18/12])12-1, = (1.015)12-1, = 0.195618..., or an APR of 19.5618%. Installment Loans An installment loan is a form of an arrangement called an annuity. In this scenario, a lender loans a borrower a principal amount P, on which the borrower will pay an interest rate of i per year for n years, with q equal payments per year, and the borrower also repays a fixed amount M to the lender q times per year. At the end of the n years, the last payment by the borrower pays off the last part of the principal and interest owed. After k payments, the amount A still owed is A = P(1+[i/q])k - Mq([1+(i/q)]k-1)/i, = (P-Mq/i)(1+[i/q])k + Mq/i. The amount of the fixed payment is determined by M = Pi/[q(1-[1+(i/q)]-nq)]. The amount of principal that can be paid off in n years is P = M(1-[1+(i/q)]-nq)q/i. The number of years needed to pay off the loan is n = -log(1-[Pi/(Mq)])/(q log[1+(i/q)]). The total amount paid by the borrower is Mnq, and the total amount of interest paid is I = Mnq - P. Example: Suppose you finance your car with a loan of $12000 at an interest rate of 11% for four years, and make equal payments monthly. How much will your payments have to be? Here the parameters are principal P = $12000, interest rate i = 0.11, number of years n = 4, and number of periods per year q = 12. Then the monthly car payment M is given by M = Pi/[q(1-[1+(i/q)]-nq)], = ($12000)(0.11)/[(12)(1-[1+(0.11/12)]-(4)(12))], = $110/(1-1.009166666...-48), = $310.15. How much will be owed at the end of two years? Here we have a monthly payment of M = $310.15, and the number of periods k = 2q = 24, and compute the amount due A at that time. A = (P-Mq/i)(1+[i/q])k + Mq/i, = ($12000-[$310.15][12]/0.11)(1+[0.11/12])24 + ($310.15)(12)/0.11, = -$27180.264935 + $33834.545454, = $6654.28. How much interest will have been paid in total? I = Mnq - P, = ($310.15)(4)(12) - $12000, = $14887.20 - $12000, = $2887.20. Example: You purchase a new home for $250,000, and pay $50,000 as a down payment. You take out a mortgage for the remaining $200,000 at 8% interest for 30 years, compounded monthly, with equal monthly payments. What is your monthly mortgage payment? Here the parameters are principal P = $200000, interest rate i = 0.08, number of years n = 30, and number of periods per year q = 12. Then your monthly payment M is given by M = Pi/[q(1-[1+(i/q)]-nq)], = ($200000)(0.08)/[(12)(1-[1+(0.08/12)]-(30)(12))], = ($1333.333333...)/(1-[1.006666666...]-360), = $1467.53. The interest rate i can't be solved for algebraically, and must be found numerically. One way to do this is as follows. Guess a value i1 for i. A reasonable guess that will be too high is the interest rate for simple interest, i1 = Mnq/P - 1. Using that, compute the principal P1 for that rate. If P1 < P, then the interest rate i1 > i, but if P1 > P, then i1 < i. Using that fact, try another interest rate i2 and compute the corresponding principal value P2. Then try the new rate, i3 = (i1[P2-P]+i2[P-P1])/(P2-P1), which is gotten by linear interpolation. Replace the worse of the two starting interest rates with this new rate i3. Repeat this, always using the two interest rates with the corresponding principals closest to P as i1 and i2. Continue until you have found an interest rate such that the corresponding principal when rounded to the nearest cent gives P. Then i can be taken equal to that interest rate. Example: What annual interest rate would you be paying if you paid off a principal of $1000 in two years with monthly payments of $50? Here P = $1000, n = 2, q = 12, and M = $50, and you want to find i. Here is a table of guesses for i and values obtained from them by interpolation, and the corresponding principal amounts. i P P - $1000 0.20 $982.40 -$17.60 0.18 $1001.52 $1.52 0.18159 $999.98 -$0.019221 0.181570145 $1000.00 -$0.000017576 Thus the interest rate would be 18.157% per year. Regular Deposits Suppose you opened an account at a bank which was paying an annual interest rate of i (a fraction, equivalent to 100i%). You make a deposit of M at the end of each of q equal time periods each year (including the end of the last period). The interest is compounded once per period. Then the value P of the account at the end of n years is given by P = M([1+(i/q)]nq-1)(q/i) This can, of course, be solved for M or n algebraically: M = iP/[q([1+(i/q)]nq-1)] n = (1/q) log(1 + [(iP)/(Mq)])/log(1 + [i/q]) The interest rate i can't be solved for algebraically, and must be found numerically. (See elsewhere on this page for methods for doing this.) Example: Suppose you deposit $100 in a bank at the end of every month, and receive 6% per year compound interest for 4 years, compounded monthly. The parameters will be payment M = $100, annual interest rate i = 0.06, number of years n = 4, and number of periods per year q = 12. Over these 4 years the principal will grow to the amount P, where P = M([1+(i/q)]nq-1)q/i = ($100)([1+(0.06/12)](4)(12)-1)12/0.06 = ($100)(1.00548-1)200 = $5409.78 Note that if you make the deposits at the start of each period, you get an extra month of interest, which you can compute by multiplying the total by (1+(i/q)): P = M([1+(i/q)]nq-1)(q/i)(1+(i/q)) = M([1+(i/q)]nq-1)(1+(q/i)) Question 1 The March, 1974 edition of _Scientific American_ said that the rate of energy consumption increased 3.1 percent per year between 1947 and 1971. If the coal reserves would last 600 years at the 1972 consumption rate, how long would they last if consumption continued to increase at the rate of 3.1 percent per year? Answer log(600(0.031) + 1)/log(1.031) =97 solve it. NOTES The time period for principal doubling itself is given by the approximation formula n=(72/r) n=69/r+0.35 r is rate of interest for one time period. The above empirical formulae would be valid for most of the common rate of interest and time periods. The error would increase if r decreases below 5%. Both the formulae give very close values

Series

Arithmetic Mean Definition:
Arithmetic mean is commonly called as average. Mean or Average is defined as the sum of all the given elements divided by the total number of elements.

Formula:
Mean = sum of elements / number of elements
= a1+a2+a3+.....+an/n

Example: To find the mean of 3,5,7
15/3 = 5

Note:

1. Mean is defined only for real positive number input. No negative numbers is allowed as input
2. The number of input is finite up to n. (not infinite)
3. The mean value is somewhere in between the lowest and the highest input numbers. The mean value will not go outside this range of input.

Geometric Mean
The geometric mean is the nth root of the product of the scores. Thus, the geometric mean of the scores: 1, 2, 3, and 10 is the
fourth root of 1 x 2 x 3 x 10 which is the fourth root of 60 which equals 2.78. The formula can be written as:

GM= (a₁*a₂*a₃*.......a_n)^1/n The geometric mean between two numbers is: √(f₁ · f₂₎

Harmonic mean Harmonic mean is the reciprocal of arithmetic mean of reciprocal.

NOTE: 1. If the data is all positive, the arithmetic mean is always greater than or equal to the geometric mean of the same data

2. The log of the geometric mean of a set of numbers is the arithmetic mean of the logs of the numbers:
log((abc)^(1/3)) = (1/3)log(abc) = (1/3)(log(a)+log(b)+log(c))

Arithmetic Series

A military unit purchases 10 spare parts during the first month of a contract, 15 spare parts in the second month, 20 spare parts in the third month, 25 spare parts in the fourth month, and so on. The acquisition officer wants to know the total number of spare parts the unit will have acquired after 50 months. This sequence of number of parts purchased in each month is called an Arithmetic Series and the sum of this series (i.e., the total number of purchased spare parts) can be written as follows.

Geometry

Basics

Angles

Polygon

Area

Space Figure

General Formulas

Triangle

Circles

Ratio & Proportion

A ratio is a comparison of two numbers. We generally separate the two numbers in the ratio with a colon ":". Suppose we want to write the ratio of 8 and 12.
We can write this as 8:12 or as a fraction 8/12, and we say the ratio is eight to twelve.

Examples :

• In a certain class, the ratio of passing grades to failing grades is 7 to 5. How many of the 36 students failed the course?

The ratio, "7 to 5" (or 7 : 5 or ⁷/₅), tells you that, of every 7 + 5 = 12 students, five failed. That is, ⁵/₁₂ of the class flunked. Then ( ⁵/₁₂ )(36) = 15 students failed.

• Express the ratio in simplest form: $10 to $45

This means that you should write the ratio as a fraction, and you should then reduce the fraction:

.¹⁰/₄₅ = ²/₉.

This reduced fraction is the ratio's expression in simplest form. Note that the units "canceled" on the fraction, since the units, "$", were the same on both values. So there is no unit on the answer.

• Express the ratio in simplest form: 240 miles to 8 gallons

Depending on the text (or instructor), you may be supposed to keep the units on a ratio. In this case, you would have (240 miles)/(8 gallons) = (30 miles)/(1 gallon), or, in more common language, 30 miles per gallon.

NOTES
1.Compounded Ratio of two ratios a/b and c/d is ac/bd, i.e., ac : bd.
Duplicate ratio of a : b is a² : b2
Triplicate ratio of a : b is a³ : b³
Sub-duplicate ratio of a : b is a : b
Sub-triplicate ratio of a : b is ³a : ³b
Reciprocal ratio of a : b is b : a

2.Proportion. Four (non-zero) quantities of the same kind a, b, c, d are in proportion,
written as a : b :: c : d iff a/b = c/d

3.The non-zero quantities of the same kind a, b, c, d, ... are in continued proportion iff
a/b = b/c = c/d = ...
In particular, a, b, c are in continued proportion iff a/b = c/d. In this case b is called the mean proportion; b = ac; c is called third proportional. If a, b, c, d are in proportion, then d is called fourth proportional.

4.Invertendo. If a : b :: c : d then b : a :: d : c
Alternendo. If a : b :: c : d then a : c :: b : d
Componendo. If a : b :: c : d then (a +b) : b :: (c +d) : d
Dividendo. If a : b :: c : d then (a -b) : b :: (c -d) : d
Componendo and dividendo.
If a : b :: c : d then (a +b) : (a -b) :: (c +d) : (c -d)
i.e., a/b = c/d => (a +b)/(a - b) = (c +d)/(c +d)

5.If a/b = c/d = e/f = ..., then each ratio = (a +c +e +...)/(b +d +f +...)
Which is same as ( pa + qc + re +. . . )/( pb + qd + rf + . . . )
i.e
   a/b = c/d = e/f
=(a +c +e +...)/(b +d +f +...)
= ( pa + qc + re +. . . )/( pb + qd + rf + . . . )
= ( pⁿa + qⁿc + rⁿe +. . . )^1/n/( pⁿb + qⁿd + rⁿf + . . . )^1/n

where p, q, r, etc. are constants such that all of them are simultaneously not equal to zero

Examples 1. If a doctor prescribes 30 units of insulin in 500 ml to be administered over 2 hours, how many drops per minute should be administered if the set is calibrated to deliver 20 drops per ml?

The rate we want is 500 milliliters over two hours, which we can write as a fraction: 500 ml / 2 hr We know that there are 20 drops per milliliter, so that 20 drops/1ml = 1 since the numerator is the same amount as the denominator. If we multiply the rate by this fraction, we won't change the rate. We then have ml in the numerator and the denominator, so they will cancel:

(500 ml /2hr)*(20 drops /1ml) = (500*20) drops / (2*1hr) = 5000 drops/hr

We're not done yet, because we need to know the number of drops per MINUTE. But that's easy using another "unit multiplier". Since 1 hour = 60 minutes, we can write 1 hour / 60min = 1

I chose to put the hour on top to cancel the hours in the bottom of our fraction, 5000 drops/hour, when we multiply:

( 5000 drops/ 1 hour) *(1hour/60 mins) = 5000/60 drops/min = 83.333 drops/min

2. A petroleum distributor has two gasohol storage tanks, the first containing 9 percent alcohol and the second containing 12 percent alcohol. They receive an order for 300,000 gallons of 10 percent alcohol. How can they mix alcohol from the two tanks to fill this order?

Let x = volume of type(1) 100-x = volume of type(2) in 100 gallons of mixture. Then 0.09x + 0.12(100-x) = 10 gallons -0.03x + 12 = 10 2 = 0.03x and so x = 66.6667 gallons So we use 66+(2/3) gallons type (1) and 33+(1/3) gallons type (2) per 100 gallons of mixture. For 300,000 gallons, we multiply up by 3000 to get 200,000 and 100,000 of type (1) and type (2) respectively.

3.A cable television company plans to begin operations in a small town. The company knows that about 600 people will subscribe to the service if the price per subscriber is $5 per month, but that for each 5-cent increase in the monthly subscription price, 4 of the original 600 people will decide not to subscribe. The office begins operations, and its total revenue for the first month is $1500. How many people have subscribed to the cable television service?

Let x = number of increments of 5-cents added to the price. So the number of subscribers = 600-4x, and subscription = 5+.05x So we have: (600-4x)(5+.05x) = 1500 3000 - 20x + 30x - 0.2x^2 = 1500 -0.2x^2 + 10x + 1500 = 0 0.2x^2 - 10x - 1500 = 0 so x = -65.1387 or x = 115.13 The negative value we ignore, as the question does not say that they would gain subscribers by reducing the price. So they have increased the price by 115 increments of 5 cents = $5.75 Total cost of subscribing = $5 + $5.75 = $10.75 Total number of subscribers = 600-460 = 140 Total income = 140*10.75 = $1505 - which checks approximately.

4.Milk that has 5% butterfat is mixed with milk that has 2% butterfat. How much of each is needed to obtain 60 gallons of milk that has 3% butterfat0.05x + 0.02y = 0.03*60 = 1.8 You might like to avoid decimals by multiplying the whole equation by 100: 5x + 2y = 180 Now you have a system of simultaneous equations: x + y = 60 5x + 2y = 180 You have probably learned some method of solving this. If not, the easiest way is to solve the first equation for y: y = 60 - x and then replace y in the second equation with (60-x). This gives you a single equation to solve to find x.

5.Sterling Silver is 92.5% pure silver. How many grams of pure silver and sterling silver must be mixed to obtain 100g of a 94% Silver alloy?

The amounts of the two input mixtures are unknown; the amount of the output mixture is 100g. Let x = grams of 92.5% silver alloy Then, since the total weight is 100g, we have (100-x) = grams of 100% silver We now write an equation relating the amounts of pure silver in the two "input" mixtures and in the "output" mixture: "x" grams at 92.5% silver + (100-x) grams at 100% = 100g at 94% (x)(0.925) + (100-x)(1.0) = (100)(0.94) 0.925x + 100 - x = 94 6 = 0.075x 6000 = 75x 6000/75 = x 80 = x So to make 100g of an alloy of 94% silver, you need to mix 80g of 92.5% alloy and 20g of pure silver.

And now here is a completely different approach to the same problem. I prefer this method, because I find the calculations are usually easier. Understanding why this method works is probably a bit more difficult than understanding the traditional method, but it works for me, so I use it. Take a look at this alternative method and see if you like it. We have two "input" mixtures; one of 92.5% silver and the other of 100% silver. We want to make a mixture of 94% silver. If I think of plotting these percentages on a number line, I see that the "distance" from 92.5% to 94% is 1.5%, while the "distance" from 94% to 100% is 6%. And now here is the key to my method: The distances from 92.5% to 94% and from 94% to 100% are 1.5% and 6%; these two distances are in the ratio 1:4. This means that the two "input" mixtures must be mixed in the ratio 1:4 to get the 94% alloy. If there are to be 100g of the 94% alloy and the two input mixtures are in the ratio 1:4, then there must be 20g of one input and 80g of the other. Because the resulting alloy is closer to 92.5% than 100%, the required amounts of the inputs are 80g of the 92.5% alloy and 20g of the 100% silver.

6.Person A has a cup of coffee. Person B has a cup of tea. Person A takes a teaspoon of coffee and puts it in the tea. Person B mixes the teaspoon of coffee and tea, then takes a teaspoon of the mixture and returns it to the coffee. Is there more coffee in the tea, or more tea in the coffee, or are they the same?

The volume of each cup after the swap is the same. That means that whatever coffee is missing from the first cup must have been replaced with exactly the same amount of tea.

7.Chris has 3 gallons of a solution that is 30 percent antifreeze, which he wants to use to winterize his car. How much pure antifreeze should he add to this solution so that the new solution will be 65 percent antifreeze?

you start out with x gallons of antifreeze that's been mixed with y gallons of water, and the total is 3 gallons, so I'd say x + y = 3 is a good equation to start with. You also know that the solution is 30 percent antifreeze, so x/(x+y) = .30. Since you have two equations and two unknowns, these equations are enough to tell you both x and y. Maybe you should find x and y before going any further, and then you can write the next equation more simply. Now we want to add some more antifreeze until the total is 65 percent antifreeze. Say you're adding z gallons. The equation is (x+z)/(x+y+z) = 0.65 The x+z is the total amount of antifreeze, and the x+y+z is the total amount of solution, antifreeze+water. Since you already know x and y, you can put them into this equation and you'll just have a single equation with z in it. That should do the trick!

8.Suppose 30 liters of a solution with an unknown percentage of alcohol is mixed with 5 liters of a 90% alcohol solution. If the resulting mixture is a 62% alcohol solution, what is the percentage of alcohol in the first solution?

Think of the amount of alcohol in each solution. In the first one, it is 30*(x/100) liters. In the second one, it is 5*(90/100). In the last one, it is 35*(62/100) (because the total amount of the mixture is 35 = 30 + 5 liters). When the first two solutions are mixed, the total amount of alcohol is the sum of the amounts in the two ingredients, so you get an equation: 30*(x/100) + 5*(90/100) = 35*(62/100).

9.A group of chemists are conducting an experiment to produce a new liquid material. One chemical contains 15% sodium (Na) and the other chemical contains 30% sodium (Na). Once they mix the two samples the resulting chemical contains 22% sodium (Na). How many milliliters (ml) of each sample must be mixed to obtain 600 ml of the new chemical?

Since the first material is 15% sodium, it means that if we have some amount, A, of the material, we can determine the amount of sodium it contains by taking 15% of that. Does that make sense? amount of sodium from first material = 0.15*A For example, if we have 100 ml of the material, then 15 ml of it will be sodium, and 85 ml will be something else. Similarly, if B is the amount of the second material, amount of sodium from second material = 0.30*B Now, if we combine the two chemicals to get a new chemical, which is 22% sodium, then the total amount of combined sodium is amount of sodium from both sources = 0.22*(A+B) So we can use that to set up an equation: sodium from sodium from sodium from first material + second materal = both materials 0.15*A + 0.30*B = 0.22*(A+B) Now, this is kind of a problem, since we have two variables but only one equation. But in fact, we _have_ a second equation, because we know that A + B = 600 So now you have two equations, and you can use substitution or elimination to find the values of the variables:

10.An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice?

You have 5 quarts of "drink" that is 10% orange juice. Thus the 5 quarts of drink contain 5*10% = 5*.1 = .5 quarts of orange juice, and hence must contain 4.5 quarts of other stuff. Now we're going to add 100% orange juice and we want to know how much to add so that the new mixture will be 40% orange juice. Well, what if you add 1 quart of 100% orange juice? What would that do to the concentration of orange juice in the mixture? It would mean that we now have 1.5 quarts of pure orange juice mixed in a carton that now contains 6 quarts of liquid. Thus the concentration of orange juice in this 6-quart jug would be 1.5/6 = .25 = 25% So we haven't added enough, but maybe you see what's going on. In a more general situation: If we add x quarts of pure orange juice to the original mixture that contains .5 quarts of OJ and 4.5 quarts of something else, then what we have is a mixture of 5+x quarts that contains .5+x quarts of OJ. So after we've added x quarts of pure OJ, the concentration of OJ will be (.5+x)/(5+x) We want to know when this will be equal to .4 So solve for x: (.5+x)/(5+x) = .4 .5+x = 2 + .4x .6x = 1.5 x = 1.5/.6 = 2.5 Thus you should add 2.5 quarts of pure OJ to make the concentration of the new mixture 40% OJ

Time , Speed and Distance

We define speed as distance divided by time, Speed = distance /time (D=ST) but once we have the equation, we can use any of its variations, speed = distance / time , distance = speed * time and time = distance / speed to compute any one of the quantities when we happen to know the other two. For example, suppose we drive for 2 hours at 30 miles per hour, for a total of 60 miles. If we know the time and the speed, we can find the distance: 2 hours * 30 miles/hour = 60 miles If we know the time and the distance, we can find the speed: 60 miles / 2 hours = 30 miles/hour Distance is directly proportional to Velocity when time is constant 1. A car travels at 30km/hr for the first 2 hrs & then 40km/h for the next 2hrs. Find the ratio of distance traveled S1/S2=V1/V2=3/4 2. Two cars leave simultaneously from points A & B (100km apart)& they meet at a point 40 km from A. What is Va/Vb? T is constant so V1/V2=S1/S2=40/60=4/6 3. A train meets with an accidient and moves at 3/4th its original speed. Due to this , it is 20 mins late. Find the original time for the journey beyond the point of accident? Method1 : Think about 2 diff. Situations , 1st with accident and another w/o accident , then S is constant So V1/V2=T1/T2 =>V1/[3/4)*V2]=(T1+20)/T1 =>4/3=(T1+20)/T1 =>T1=60 Method 2: Velocity decreases by 25% so time will increase by 33.3% 33.3%=20 mins =>100%=60 mins CONVERSION: 1km/hr=1000m/h=1000/3600m/sec=5/18m/sec

RELATIVE SPEED

Set Theory

Naive set theory considers elementary properties of the union and intersection operators -- Venn diagrams, the DeMorgan laws, elementary counting techniques such as the inclusion-exclusion principle, partially ordered sets, and so on. This is perhaps as much of set theory as the typical mathematician uses. Indeed, one may "construct" the natural numbers, real numbers, and so on in this framework.

1. Ordered Pairs

We begin by introducing the notion of the ordered pair. If a and b are sets, then the unordered pair {a, b} is a set whose elements are exactly a and b. The “order” in which a and b are put together plays no role; {a, b} = {b, a}. For many applications, we need to pair a and b in a way making possible to “read off” which set comes “first” and which comes “second.” We denote this ordered pair of a and b by (a, b); a is the first coordinate of the pair (a, b), b is the second coordinate.

As any object of our study, the ordered pair has to be a set. It should be defined in such a way that two ordered pairs are equal if and only if their first coordinates are equal and their second coordinates are equal. This guarantees in particular that (a, b) ≠ (b,a) if a ≠ b.

Definition. (a, b) = {{a}, {a, b}}.

2. Relations

A binary relation is determined by specifying all ordered pairs of objects in that relation; it does not matter by what property the set of these ordered pairs is described. We are led to the following definition.

Definition. A set R is a binary relation if all elements of R are ordered pairs, i.e., if for any z ∈ R there exist x and y such that z = (x, y).

It is customary to write xRy instead of (x, y) ∈ R. We say that x is in relation R with y if xRy holds.

The set of all x which are in relation R with some y is called the domain of R and denoted by “dom R.” So dom R = {x | there exists y such that xRy}. dom R is the set of all first coordinates of ordered pairs in R.

The set of all y such that, for some x, x is in relation R with y is called the range of R, denoted by “ran R.” So ran R = {y | there exists x such that xRy}.

3. Functions

Function, as understood in mathematics, is a procedure, a rule, assigning to any object a from the domain of the function a unique object b, the value of the function at a. A function, therefore, represents a special type of relation, a relation where every object a from the domain is related to precisely one object in the range, namely, to the value of the function at a.

4. Natural Numbers

In order to develop mathematics within the framework of the axiomatic set theory, it is necessary to define natural numbers. We all know natural numbers intuitively: 0, 1, 2, 3, ..., 17, …, 324, etc., and we can easily give examples of sets having zero, one, two, or three elements.

To define number 0, we choose a representative of all sets having no elements. But this is easy, since there is only one such set. We define 0 = Ø. Let us proceed to sets having one element (singletons): {Ø}, {{Ø}}, {{Ø, {Ø}}}; in general, {x}. How should we choose a representative? Since we already defined one particular object, namely 0, a natural choice is {0}. So we define

1 = {0} = {Ø}.

2 = {0,1} = {Ø, {Ø}}.

5. Cardinality of Sets

From the point of view of pure set theory, the most basic question about a set is: How many elements does it have? It is a fundamental observation that we can define the statement “sets A and B have the same number of elements” without knowing anything about numbers.

Definition. Sets A and B have the same cardinality if there is a one-to-one function f with domain A and range B. We denote this by |A| = |B|.

Definition. The cardinality of A is less than or equal to the cardinality of B (notation: |A| ≤ |B|) if there is a one-to-one mapping of A into B.

For example the cardinality of the set {3, 1, 2} is 3.

6. Finite Sets

Finite sets can be defined as those sets whose size is a natural number.

Definition. A set S is finite if it has the same cardinality as some natural number n ∈ N. We then define |S| = n and say that S has n elements. A set is infinite if it is not finite.

7. Countable Sets

Definition. A set S is countable if |S| = |N|. A set S is at most countable if |S| ≤ |N|.

Thus a set S is countable if there is a one-to-one mapping of N onto S, that is, if S is the range of an infinite one-to-one sequence.

8. Uncountable Sets :

All infinite sets whose cardinalities we have determined up to this point turned out to be countable. Naturally, a question arises whether perhaps all infinite sets are countable. If it were so, this book might end with the preceding section. It was a great discovery of Georg Cantor that uncountable sets, in fact, exist. This discovery provided an impetus for the development of set theory and became a source of its depth and richness.

9.Null Set:
In mathematical sets, the null set, also called the empty set, is the set that does not contain anything. It is symbolized for { }. There is only one null set. This is because there is logically only one way that a set can contain nothing.

The null set makes it possible to explicitly define the results of operations on certain sets that would otherwise not be explicitly definable. The intersection of two disjoint sets (two sets that contain no elements in common) is the null set. For example:

{1, 3, 5, 7, 9...} Ç{2, 4, 6, 8, and 10...} = f

10.Equality of sets: Two sets are equal if and only if they have the same elements.
More formally, for any sets A and B, A = B if and only if all the elements of A present in B.

Thus for example {1, 2, 3} = {3, 2, 1}, that is the order of elements does not matter, and {1, 2, 3} = {3, 2, 1, 1}, that is duplications do not make any difference for sets.

11.Subset: A set A is a subset of a set B if and only if everything in A is also in B.
More formally, for any sets A and B, A is a subset of B, and denoted by A ÍB, if and only if all the elements of A are present in B
If A ÍB, and A ¹B, then A is said to be a proper subset of B and it is denoted by A ÌB .

For example {1, 2} Í{3, 2, 1}. Also {1, 2}Ì{3, 2, 1}.

12.Universal set: A set, which has all the elements in the universe of discourse, is called a universal set.

Three subset relationships involving empty set and universal set are listed below as theorems without proof.

Note that the set A in the next four theorems are arbitrary. So A can be an empty set or universal set.

Theorem 1: For an arbitrary set A: A ÍU.

Theorem 2: For an arbitrary set A: fÍA.

Theorem 3: For an arbitrary set A A ÍA.

13.Power set: The set of all subsets of a set A is called the power set of A and denoted by 2^A or j(A) .

For example for A = {1, 2}, j (A) = {f, {1}, {2}, {1, 2} } .

Theorem 4: For an arbitrary set A, the number of subsets of A is 2^|A|.

Operation of sets II

EXERCISE

Let the universal set be the set of all people. Give the complement of the set of all people who can swim.

Complement of a Complement

The complement of a complement is written in symbols as follows: (A')' = {x: not-(x is not in A)}.

Since not-(x is not in A) has the same meaning as x is in A, it follows that (A')' = {x: x is in A}. But {x: x is in A} = A. Therefore we have the complement law:

(A')' = A.

For example, let the universal set be the set of whole numbers from 1 to 10, and let A be the set of even numbers from 2 to 10. Then A' is the set of odd numbers from 1 to 9, and (A')' is the original set A of even numbers from 2 to 10.

EXERCISE

Let the universal set consist of all the books in a library. Let A be the set of books written in English. What are the sets A' and (A')'?

Union of a Set and its Complement

The union of a set and its complement is the universal set.

For example, let x stand for an animal, let A = {x: x is male}, and let A' = {x: x is female}. Then A + A' = {x: x is male or x is female} = the set of all animals.

We write this law briefly as follows:

A È A' = U.

EXERCISE

Give the universal sets for the following sets and their complements:

1. A = all buildings with only one floor.
   A' = all buildings with two or more floors.
2. A = all students who passed an examination
   A' = all students who failed the examination.

Intersection of a Set and its Complement

The intersection of a set and its complement is the empty set.

For example, let A consist of all the unbroken plates in a set of plates, and let A' consist of all the broken plates. Then A . A' is empty because there are no plates that are unbroken and broken at the same time.

We summarize this law in symbols as follows:

A Ç A' = O.

EXERCISE

Say whether or not the intersection of the following pairs of sets is the empty set:

1. A = all sunny days in a year
   B = all windy days in the year.
2. A = all wet days in a year
   B = all dry days in the year.

Complement of a Union

Since the union of two sets A and B is given by

A + B = {x: x is in A or x is in B},

the complement of the union is given by

(A + B)' = {x: not-(x is in A or x is in B)}

But not-(x is in A or x is in B) has the same meaning as x is not in A and x is not in B, and

{x: x is not in A and x is not in B} = A'.B'.

Therefore, we have the following law for the complement of a union:

(A È B)' = A' Ç B'.

This is one of De Morgan's laws for sets.

For example, let the universal set be the set of all substances. Let A be the set of all solids, such as stone and iron. Let B be the set of all liquids, such as water and oil. Then A + B is the set all substances that are solid or liquid. The complement (A + B)' is the set of all substances that are not solid or liquid, in other words the set A' . B' of all substances that are not solid and not liquid, such as oxygen and nitrogen (which are gases).

EXERCISE

at A = all women and girls, let B = all children (girls and boys). Say what people are members of the complement of A + B.

Complement of an Intersection

Since the intersection of two sets A and B is given by

A . B = {x: x is in A and x is in B},

the complement of this intersection is given by

(A . B)' = {x: not-(x is in A and x is in B)}.

But not-(x is in A and x is in B) has the same meaning as x is not in A or x is not in B, and

{x: x is not in A or x is not in B} = A' + B'.

Therefore, we have the following law for the complement of an intersection:

(A Ç B)' = A' È B'.

This is another of De Morgan's laws for sets.

For example, let the universal set be the whole numbers from 1 to 10. Let A = {1, 2, 3, 4, 7, 8, 9, 10} and let B = {2, 4, 6, 8, 10}. Then A . B = {2, 4, 8, 10}, and so (A . B)' = {1, 3, 5, 6, 7, 9}. Also A' = {5, 6}, B' = {1, 3, 5, 7, 9} and A' + B' = {1, 3, 5, 6, 7, 9}. Therefore, in this example, (A . B)' = A' + B'.

EXERCISE

Let A = all women and girls, and let B = all children (girls and boys). Say what people are members of the complement of A . B when the universal set is all people.

Operation of sets I

Union of Sets

From two given sets A and B we can make a new set that consists of all the elements of A and all the elements of B. This new set is called the union of A and B. It is represented by the symbol A + B.

For example, let A = {a, b, c, d} and let B = {c, d, e}. Then A + B = {a, b, c, d, e}. Notice that the elements c and d are in A as well as B, but they are written only once in the list for A È B.

The union of two sets is defined in symbols as follows:

A È B = {x: x is in A or x is in B}.

Intersection of Sets

From two given sets A and B we can make a new set that consists of all the elements that belong to both A and B at the same time. This new set is called the intersection of A and B. It is represented by the symbol AÇB.

For example, let A = {a, b, c, d} and let B = {c, d, e}. Then A Ç B = {c, d}.

The intersection of two sets is defined in symbols as follows:

A Ç B = {x: x is in A and x is in B}.

EXERCISE

Give the intersections of the following pairs of sets.

A = {2, 4, 6, 8, 10, 12}
B = {3, 6, 9, 12}.

A = the time from 6:00 a.m. to 6:00 p.m.
B = the time from 12:00 noon to 12:00 midnight

Commutative Law for UnionsIn the union of two sets it does not matter which set is written first. For example,

{a, b, c} È {c, d} = {a, b, c, d}

and

{c, d} È {a, b, c} = {a, b, c, d}.

This may also be seen in the definitions

A È B = {x: x is in A or x is in B}

and

B È A = {x: x is in B or x is in A}

because, by the commutative law for the logical OR, x is in A or x is in B has the same meaning as x is in B or x is in A.

This fact is called the commutative law for the union of sets. It is summarized in symbols as follows:

A È B = B È A.

Associative Law for Unions

The union of A È B with another set C is a composite union:

(A + B) + C = {x: (x is in A or x is in B) or x is in C}.

By the associative law for the logical OR, this is equal to:

{x: x is in A or (x is in B or x is in C)} = A + (B + C).

Therefore we have the associative law for unions, which is summarized as follows:

(A È B) È C = A È (B È C).

As a result of this law we may omit the brackets and write:

A + B + C.

For example, let A be the set of all African people, let B be the set of all Asian people, and let C be the set of all European people. Then the union A + B + C is the set of all the people in Africa, Asia, and Europe.

Commutative Law for Intersections

In the intersection of two sets it does not matter which set is written first. For example,

{a, b, c, d} Ç {c, d, e} = {c, d}

and

{c, d, e} Ç {a, b, c, d} = {c, d}.

This may also be seen in the definitions

A Ç B = {x: x is in A and x is in B}

and

B Ç A = {x: x is in B and x is in A}

because, by the commutative law for the logical AND, x is in A and x is in B has the same meaning as x is in B and x is in A.This fact is called the commutative law for the intersection of two sets. It is summarized in symbols as follows:

A Ç B = B Ç A.

Associative Law for Intersections

The intersection of A Ç B with another set C is a composite intersection:

(A Ç B) Ç C = {x: (x is in A and x is in B) and x is in C}.

By the associative law for the logical AND, this is equal to

{x: x is in A and (x is in B and x is in C)} = A Ç (B Ç C).

Therefore we have the associative law for intersections, which is summarized in symbols as follows:

(A Ç B) Ç C = A Ç (B Ç C)

As a result of this law we may omit the brackets and write

A Ç B Ç C

For example, let A be the set of all African people, let B be the set of all female people, and let C be the set of all children under 10 years old. Then A . B . C is the set of all African girls under 10 years old.

Distributive Law: Union over Intersection

The union of a set A with an intersection B . C is a composite expression:

A + (B . C) = {x: x is in A or (x is in B and x is in C)}.

By one of the distributive laws for composite statements, x is in A or (x is in B and x is in C) has the same meaning as (x is in A or x is in B) and (x is in A or x is in C). Therefore

A + (B . C) = {x: (x is in A or x is in B) and (x is in A or x is in C)}.

But

{x: (x is in A or x is in B) and (x is in A or x is in C)} = (A + B) . (A + C).

Therefore

A È (BÇC) = (A È B) Ç (A È C).

This is one of the distributive laws for sets.

For example, suppose that applicants for a job must have either a university degree, or five years of work experience and a certificate of English language ability. Let A be the set of people with a degree, let B be the set of people with five years of experience, and let C be the set of people with a certificate of English. Then A + (B . C) is the set of people who may apply for the job. By the distributive law this is the same as (A + B) . (A + C). This shows that the applications may be checked in two separate ways before being accepted. One check makes sure that the applicant belongs to A + B (has a degree or five years of experience). The other check makes sure that the applicant belongs to A + C (has a degree or a certificate of English).

EXERCISE

Suppose a person has enough money to buy a small new motor-car or a large second-hand motor-car. Let A be the set of small cars, let B be the set of large cars, and let C be the set of second-hand cars. Use the distributive law to find another way of saying that this person may buy a car which is small or large, and small or second-hand.

Distributive Law: Intersection over Union

The intersection of a set A with a union B + C is a composite expression:

A . (B + C) = {x: x is in A and (x is in B or x is in C)}.

By one of the distributive laws for composite statements, x is in A and (x is in B or x is in C) has the same meaning as (x is in A and x is in B) or (x is in A and x is in C). Therefore

A . (B + C) = {x: (x is in A and x is in B) or (x is in A and x is in C)}.

But {x: (x is in A and x is in B) or (x is in A and x is in C)} = (A . B) + (A . C).

Therefore

A Ç (B È C) = (A Ç B) È (A Ç C).

This is another distributive law for sets.

For example, let A be the set of children from 5 to 12 years old, let B be the set of boys, and let C be the set of girls. Then A . (B + C) means: the set of children who are 5 to 12 years old and either boys or girls. This is the same as the set (A . B) + (A . C), which means: children who are boys 5 to 12 years old or girls 5 to 12 years old.

EXERCISE

Let A be the set of windy days in a particular year at a particular place, let B be the set of sunny days, and let C be the set of rainy days. Give the meaning of the two composite expressions:

A . (B + C)

(A . B) + (A . C).

Complement

The complement of a set A is the set of all elements in the universal set that do not belong to A. The complement of A is represented by the symbol A'. It is defined in symbols as follows:

A' = {x: x is not in A}.

For example, let the universal set be the set of whole numbers from 1 to 10, and let A be the set of all even numbers from 2 to 10. That is:

A = {2, 4, 6, 8, 10}.

Then A' is the set of all odd numbers from 1 to 9:

Venn Diagrams

Interactive Diagrams
Venn diagrams are illustrations used in the branch of mathematics known as set theory. They are used to show the mathematical or logical relationship between different groups of things (sets). A Venn diagram shows all the logical relations between the sets

Permutations

BY THE SYMBOL n! ("n factorial") we mean the product of consecutive numbers 1 through n. n! = 1 x 2 x 3 x . . . . x (n - 2) x (n - 1) x n
=n x (n - 1) x (n - 2) x . . . .x 3 x 2 x 1 The order of the factors does not matter, whether backwards or forwards. 0! is defined as 1. 0! = 1 (The usefulness of this definition will become clear as we continue.) Example. 6! = 1 x 2 x 3 x 4 x 5 x 6 = 720 Any factorial less than n! is a factor of n!. Counting Principles
Counting principles describe the total number of possibilities or choices for certain selections. The two fundamental counting principles are listed below. Fundamental Counting Principle 1: If the number of events be n, and the number of outcomes for each event in an experiment be Ti (such that i = 1 for the first event, 2 for the second event, ..., and n for the nth event), then the total number of outcomes for all event is T1 x T2 x . . . . x Tn. Example: A movie theater sells 3 sizes of popcorn (small, medium, and large) with 3 choices of toppings (no butter, butter, extra butter). How many possible ways can a bag of popcorn be purchased? Answer: 3 x 3 = 9 Fundamental Counting Principle 2: If the number of mutually exclusive (no common elements) experiments be m, and the total number of outcomes for all events in each experiment be xj (such that j = 1 for the first experiment, 2 for the second experiment, ..., and m for the mth experiment), then the total number of outcomes for all experiments is x1 + x2 + ... + xm.
Arrangements Sets (n Permutations m)
Permutation is the selection of subsets from a set of elements when the order of the selected elements is a factor. BY THE PERMUTATIONS of the letters abc we mean all of their possible arrangements: abc acb bac bca cab cba There are 6 permutations of three different things The arrangements of n elements taken m at a time represent, in fact, a partial permutation. Older books use the notation nPm (n Permutation m) for arrangements: nPm = n! / (n-m)! The number of permutations of n different things taken n at a time is n!.
Important Notes: a) Number of permutations of n things, taken r at a time, when a particular thing is to be always included in each arrangement = r^n-1Pr-1 (b) Number of permutations of n things, taken r at a time, when a particular thing is fixed: = n-1 Pr-1 (c) Number of permutations of n things, taken r at a time, when a particular thing is never taken: = n-1 Pr. (d) Number of permutations of n things, taken r at a time, when m specified things always come together = m! x ( n-m+1) ! (e) Number of permutations of n things, taken all at a time, when m specified things always come together = n ! - [ m! x (n-m+1)! ] (f) The number of permutations of n elements with n1 of repeated element, n2 of another repeated element, etc. taken n at a time. P=n!/(n1!)(n2!) (g) The number of arrangement of n different things, taken r at a time, when each may be repeated any number of times in each arrangement, is n^r
Circular permutations There are two cases of circular-permutations: - (a) If clockwise and anti clock-wise orders are different, then total number of circular-permutations is given by (n-1)! (b) If clock-wise and anti-clock-wise orders are taken as not different, then total number of circular-permutations is given by (n-1)!/2! Number of circular-permutations of n different things taken r at a time:- (a) If clock-wise and anti-clockwise orders are taken as different, then total number of circular-permutations = nPr /r (b) If clock-wise and anti-clockwise orders are taken as not different, then total number of circular permutation = nPr/2r

Examples

Combinations

Combination Versus Permutation

Combination and permutation are terms that refer to the selection of a subset of objects from a larger set or pool of objects.

Combination refers to the selection process where the order of the selected objects is not important to the problem being addressed. An example of a combination is finding how many groups of four people can be formed from a population of fifteen people.

Permutation refers to the selection of a group of elements or objects from the total available number of elements or objects, and the arrangement of the selected items in a certain order. To follow the above example, if the groups of four people were sub-divided into president, vice-president, secretary and treasurer, then the order of those selected would be important and would need to be factored in when selecting the groups.

Statistical theory, in particular those concepts termed set theory, are important to computer related privacy and security issues. By use of programs that can rapidly project combinations and permutations of letters and symbols, hackers are often able to gain unauthorized access to programs and data. Such brute-force hacking often involves the generation of millions of expressions to be checked against the actual password. Although not elegant, programs without some non-randomized security element or key are often easily cracked by the high-speed generation of combinations and permutations of symbols.

Formula #1: Find the number of combinations of n elements taken r at a time.

nCr= n!/r!(n-r)!

Example:
In how many ways can a sample of 4 chocolates be selected from a box of 12 chocolates?
Answer:
₁₂C₄ = 12!/[4!—(12-4)!] = 12!/(4!—8!) = 495
Hence, there are 495 possible ways to select 12 chocolates taken 4 chocolates at a time.

Formula #2: Find the number of combinations of n elements with n₁ elements taken r₁ at a time, n₂ elements taken r₂ at a time, etc.

C = n₁Cr₁ — n₂Cr₂ — ...

(a) Number of combinations of n different things taken r at a time, when p particular things are always included = ^n-pC_r-p.

(b) Number of combination of n different things, taken r at a time, when p particular things are always to be excluded = ^n-pC_r Example: In how many ways can a cricket-eleven be chosen out of 15 players? If (i) A particular player is always chosen,(ii) A particular is never chosen.

Ans: (i) A particular player is always chosen, it means that 10 players are selected out of the remaining 14 players. =. Required number of ways = 14C10 = ¹⁴C₄= 14!/4!x19! = 1365

(ii) A particular player is never chosen, it means that 11 players are selected out of 14 players. => Required number of ways = 14C11 = 14!/11!x3! = 364

(c) Number of ways of selecting zero or more things from n different things is given by:- 2ⁿ-1

Example: John has 8 friends. In how many ways can he invite one or more of them to dinner? Ans. John can select one or more than one of his 8 friends.=> Required number of ways = 2⁸ -1 1= 255.

(d) Number of ways of selecting zero or more things from n identical things is given by :- n+1

Example: In how many ways, can zero or more letters be selected form the letters AAAAA?

Ans. Number of ways of : Selecting zero 'A's=1, Selecting one 'A's = 1 , Selecting two 'A's =1, Selecting three 'A's = 1 Selecting four 'A's = 1 , Selecting five 'A's = 1 => Required number of ways = 6 [5+1]

(e) Number of ways of selecting one or more things from p identical things of one type qidentical things of another type, r identical things of the third type and n different things is given by: - (p+1) (q+1) (r+1)2ⁿ - 1

Example: Find the number of different choices that can be made from 3 apples, 4 bananas and 5 mangoes, if at least one fruit is to be chosen.

Ans: Number of ways of selecting apples = (3+1) = 4 ways Number of ways of selecting bananas = (4+1) = 5 ways. Number of ways of selecting mangoes = (5+1) = 6 ways. Total number of ways of selecting fruits = 4 x 5 x 6 But this includes, when no fruits i.e. zero fruits is selected => Number of ways of selecting at least one fruit = (4x5x6) -1 = 119 Note: - There was no fruit of a different type, hence here n=o => 2ⁿ = 2⁰=1

(f) Number of ways of selecting r things from n identical things is 1.

(g) ⁿC_k = ⁿC_{n - k}

Probability

An experiment is a situation involving chance or probability that leads to results called outcomes.

An outcome is the result of a single trial of an experiment.

An event is one or more outcomes of an experiment.

Probability is the measure of how likely an event is.Probability starts with logic. There is a set of N elements. We can define a sub-set of n favorable elements, where n is less than or equal to N. Probability is defined as the rapport of the favorable cases over total cases, or calculated as:

P=n/N

It is the Fundamental Formula of Probability (FFPr) and everything in theory of probability is derived from it.

NOTESAn urn contains w white balls and b black balls (w > 0 and b > 0). The balls are thoroughly mixed and two are drawn, one after the other, without replacement. Let W_i and B_i denote the respective outcomes 'white on the ith draw' and 'black on the ith draw,' for i = 1, 2.

P(W2) = P(W₁) = w/(w + b). (Which clearly implies a similar identity for B₂ and B₁.)

Furthermore, P(W_i) = w/(w + b), for any i not exceeding the total number of balls w + b.

In order to measure probabilities, mathematicians have devised the following formula for finding the probability of an event.

Combining Events If E and F are events In an experiment, then:

E' is the event that E does not occur.

E U F is the event that either E occurs or F occurs (or both).

E ∩ F is the event that both E and F occur.

E and F are said to be disjoint or mutually exclusive if (E ∩F) is empty

Some Properties of Estimated Probability

Let S = {s₁, s₂, ... , s_n} be a sample space and let P(s_i) be the estimated probability of the event {s_i}. Then

(a) 0 ≤ P(s_i) ≤ 1
(b) P(s₁) + P(s₂) + ... + P(s_n) = 1
(c) If E = {e₁, e₂, ..., e_r}, then P(E) = P(e₁) + P(e₂) + ... + P(e_r).

In words:(a) The estimated probability of each outcome is a number between 0 and 1.
(b) The estimated probabilities of all the outcomes add up to 1.
(c) The estimated probability of an event E is the sum of the estimated probabilities of the individual outcomes in E.

Empirical Probability

The empirical probability, P(E), of an event E is a probability determined from the nature of the experiment rather than through actual experimentation.

The estimated probability approaches the empirical probability as the number of trials gets larger and larger.

Notes

1. We write P(E) for both estimated and empirical probability. Which one we are referring to should always be clear from the context.
2. Empirical probability satisfies the same properties (shown above) as estimated probability:

Abstract Probability

An abstract finite sample space is just a finite set S. An (abstract) probability distribution is an assignment of a number P(s_i) to each outcome s_i in a sample space S ={s₁, s₂, ... , s_n} so that

(a) 0 ≤ P(s_i) ≤ 1
(b) P(s₁) + P(s₂) + ... + P(s_n) = 1.

P(s_i) is called the (abstract) probability of s_i. Given a probability distribution, we obtain the probability of an event E by adding up the probabilities of the outcomes in E.

If P(E) = 0, we call E an impossible event. The event f is always impossible, since something must happen.

Notes 1. Abstract probability includes both estimated and empirical probability. Thus, all properties of abstract probability are also properties of estimated and empirical probability. As a consequence, everything we say about abstract probability applies equally well to estimated and empirical probability.
2. From now on, we will speak only of "probability," meaning abstract probability, thus covering both estimated and empirical probability, depending on the context.

Addition Principle Mutually Exclusive Events
If E and F are mutually exclusive events, then P(E ∩ F) = P(E) + P(F).

This holds true also for more events: If E₁, E₂, . . . , E_n are mutually exclusive events (that is, the intersection of any pair of them is empty) and E is the union of E₁, E₂, . . . , E_n, then

P(E) = P(E₁) + P(E₂) + . . . + P(E_n).

General Addition Principle
If E and F are any two events, then P(E ∩ F) = P(E) + P(F) - P(E U F).

Further Properties of Probability . /The following are true for any sample space S and any event E.

Conditional Probability

If E and F are two events, then the conditional probability, P(E|F), is the probability that E occurs, given that F occurs, and is defined by