Probability (Marbles)
Hello all, Please help me solving this problem. A box contains fewer than 10 marbles. If you reach to the box and randomly pull out 2 marbles one after the other, without replacing them you have a 50 percent chance of getting 2 blue marbles. How many blue marbles are there in the box? Thanks in Advance.
hii...
cn ya plz elabrote it mre....??? didnt get ya?
shouldn't the question
i would have done this :
Let the total number ofmarbles be x < 10
Let the total number of blue marbles be n <= x
thus the probability is
n/x * (n-1)/(x-1) = n(n-1)/x(x-1) = 1/2
we have to find a possible solution which i believe from the nature of the question only 1 exists.
no both the numerator an denominator are even.
also the denomnator has an extra 2 as factor.
since the final numerator is 1 , the numerator and denominator must have a prime number as common or they cannot have a prime factor
one solution i found was n = 3 , x =4
thus 3*2/4*3 = 1/2
Thus the number of blue marbles is 3.
lets start with x marbles
lets start with x marbles out of which n are blue.
so the probability of drawing a blue marble the first time would be n/x
now the problem states that we DONOT replace the marbles.So after the first draw we are left with x-1 marbles out of which n-1 are blue.
Thus the probability of drawing a blue marble would be n-1/x-1.
Thus the total probability of drawing 2 blue marbles in the 1st 2 trials would be n/x * (n-1)/(x-1) = n(n-1)/x(x-1) which is given as 1/2
Both the numerator and denominator are even.
x is given to be < 10 . After this follow the the rest as above or you can do it any other way. There sould be lots of them.
Solution:
Let there be x marles out of which y are blue marbles.
x < 10 , x > y
Now, the number of ways in which two marbles can be taken out without replacing =
(x C 1) * (x-1 C 1) = x(x-1).
The number of ways in which two blue marbles can be taken out without replacing =
(y C 1) * (y-1 C 1) = y(y-1).
The probability of taking out blue marbles is 1/2.
Therefore, x(x-1)/y(y-1) = 1/2
i.e. 2y(y-1) = x(x-1)
The only solution ( for x < 10 ) is x = 4, y = 3.
Thus , the no. of blue marbles is 3.