White Balls, Black Balls

Probability Level 2

There are two bags $A$ and $B$ . Bag $A$ contains $n$ white balls and $2$ black balls while bag $B$ contains $2$ white balls and $n$ black balls . One of the two bags is selected at random and two balls are drawn from it without replacement . We know that if both the balls drawn are white balls, the probability of bag $A$ being selected to draw the balls is $\frac{6}{7}$ . Find the value of $n$ .

The answer is 4.

2 solutions

Miko Co
Sep 27, 2014

The probability of choosing 2 white balls from Bag A is:

$\frac {1}{2}$ $(\frac {n}{n+2})$ $(\frac {n-1}{n+1})$ $=$ $\frac {n(n-1)}{2(n+1)(n+2)}$

The probability of choosing 2 white balls from Bag B is:

$\frac {1}{2}$ $(\frac {2}{n+2})$ $(\frac {1}{n+1})$ $=$ $\frac {1}{(n+1)(n+2)}$

The probability of selecting Bag A from the 2 bags is:

$\frac {Bag A}{Bag A+Bag B} = \frac {6}{7}$

By substituting, we get:

$\frac {\frac {n(n-1)}{2(n+1)(n+2)}}{\frac {n(n-1)}{2(n+1)(n+2)} + \frac {1}{(n+1)(n+2)}} = \frac {6}{7}$

After cross-multiplying and transposing, we get:

$\frac {n(n-1)}{2(n+1)(n+2)} = \frac {6}{(n+1)(n+2)}$

Multiplying both sides by $(n+1)(n+2)$ since $n$ is $nonnegative$ ,

$\frac {n(n-1)}{2} = 6$

$n(n-1) = 12$

$n^2 - n - 12 = 0$

$(n-4)(n+3) = 0$

Since $n$ is $nonnegative$ ,

$n-4 = 0$

Therefore, $\boxed {n=4}$

Probability of selecting bag A and selecting 2 white balls from bag A are two different things. First you select the bag then two whites.I think the question is incomplete.

Vishal Yadav - 4 years, 3 months ago

Nikhil Saxena
Sep 26, 2014

let E1 denotes the event that bag A is selected and E2 denotes the event that bag B is selected .Let E be the event that two balls drawn are white . We have

P(E1)=1/2=P(E2) P(E|E1)=n(n-1)/(n+2)(n+1) and P(E|E2)=2/(n+2)(n+1)

Using Baye's Theorem ,The probability that the two white balls drawn are from the bag A ,is given by:

P(E1|E)=(P(E1) P(E|E1))/(P(E1) P(E|E1)+P(E2)*P(E|E2))=6/7

           =n(n-1)/(n(n-1)+2)=6/7
            => n=4,-3

           since n cannot be negative ,we get

            =>  N=4

I started to write a solution but I accidentally hit "keep private" so now I can't edit it or anything bc of dumbness.

Anyways, I'll write it here.

Let us first consider the probability of choosing 2 whites from each bag.

Bag A:

2B and nW

The probability of choosing two whites is therefore $\frac{\dbinom{n}{2}}{\dbinom{n+2}{2}}$

Bag B:

nB and 2W

The probability of choosing two whites is therefore $\frac{\dbinom{2}{2}}{\dbinom{n+2}{2}}$

If you chose two whites, the probability that you chose bag A is $\frac{6}{7}$ , and then clearly the probability that you chose bag B is $\frac{1}{7}$ . The probability of choosing A is therefore 6 times that of choosing B.

Therefore,

$\frac{\dbinom{n}{2}}{\dbinom{n+2}{2}}=6 \times \frac{\dbinom{2}{2}}{\dbinom{n+2}{2}}$

$\dbinom{n}{2}=6$

$\frac{n^2-n}{2}=6$

$n^2-n-12=0$

$n = 4$ or $n=-3$

$-3<0$ therefore is inadmissable.

And so $\boxed{n=4}$

Nicolas Bryenton - 6 years, 8 months ago

White Balls, Black Balls

The answer is 4.

2 solutions

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