## quant doubts from CL

Please give me the solution of following questions?
2. The smallest positive integer X with 24 divisors is
a. 480
b. 420
c. 864
d. None of these
3. 5 coins are tossed. If two of them show heads, then the probability that all 5 coins show head is
a. 1/32
b. 1/10
c. 1/26
d. 1/13

## i have good reason to

i have good reason to believe the answer is 1/26
lets see how:
1/32 is the probability of getting all five heads when there are nor restrictions on the sample space, meaning its the worst probability!!!
getting 2 heads for given can only improve this probability. but how much ? that is the question.
The answer is pretty simple as ican see.
the number of ways of getting all 5 heads is always 1 no matter how many heads are given, which in this case is two.
the number of ways of getting at least 2 heads = total number of ways - ways to get exactly 0 heads - ways to get exactly 1 head
= 32 - 1 - 5 = 26
hence the answer 1/26

## Here it goes

Q1. Please chk the questions - is it correct?
Q2. a. 480 = 2^{4} x 3 x 5
So number of divisors = (5) x (2) x (2) = 20
b. 420 = 2^{2} x 3 x 5 x 7
So number of divisors = (3) x (2) x (2) x (2) = 24
Ans: b
Q3. I think all the events are independent
Each coin ‘ll be head/tail irrespective of others
So the probability of each event is1/2
Now the probability of 3 independents events to occur shd be 1/2 x 1/2 x 1/2 = 1/8
But its not in the option –:((

## ambiguous...

yepss....for ques no3 .....i also found da sme ansr...bt its nt in da option...if its real tym cat i wud hv gone for 1/32...
plz lemme noe..if nebuddy finds da correct ansr..
THANX..

## ritee,,

questn no 2 z CMPLETELY RIGHT....!!!!
gud wrk...

## Why so ? Why 32When 2 heads

Why so ? Why 32
When 2 heads are confirm the sample space has only 8 values.
So there is no question of 32 being in denominator
But I must be missing something :-(

## yeps buddy...u r absolutely

yeps buddy...u r absolutely rite...bt v nt hving 8 in da options...da only othr way I can tke da questn iz takin 5 heads..so v'l b gettin..1/2^5=1/32...
bt um sure...dat may b somethn z hidden in dis questn...

## Great rajorshi !!!

Great rajorshi !!!
and really thanks
I was screwing my mind for the whole day and also discussed it with the faculties of ....(well! i cant reveal anything more)

## wat the ......... your

wat the ......... your faculty ?

## Adding to rajorshi's solution

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
P(E|F) = P(E ∩ F)/P(F)
Here P(F) = Probability of getting all heads = 1/32
P(E ∩ F) = Probability of getting 2 heads and 5 heads (confusing !!)
Its like you have two head for sure and you need complete 5 heads(3 head/4 head also have two head but that wont satisfy the condn)
So the probability P(E ∩ F)= C(5,2)/32 + C(5,3)/32 + C(5,4)/32 + C(5,5)/32
=26/32
Thus P(E|F) = 1/26

## One more similar question

Try to solve the following question to understand the concept of conditional probability
A bag contains red and white cricket balls. The probability of selecting a red ball and then a white ball is 0.42, and the probability of selecting a red ball on the first draw is 0.56. What is the probability of selecting a white ball on the second draw, given that the first ball drawn was red? (Note: Balls are chosen without replacement. )