An urn contains 5 white and 5 black balls. 4 balls are drawn from this urn and put into another urn. From this second urn a ball is drawn and is found to be white. What's the probability of drawing a white ball again at the next draw. (The first white ball drawn is not replaced.)

What I did was:-

E0= Drawing 0 white and 4 black balls

E1= Drawing 1 white and 3 black balls

E2= Drawing 2 white and 2 black balls

E3= Drawing 3 white and 1 black balls

E4= Drawing 4 white and 0 black balls

We ignore E0 because if there were 0 white balls transferred, the first one drawn couldn't be white. Out of the 4 remaining events, each of them have the probability of occurring at 1/4.

In E1, if the first ball drawn is white, there are 0 white balls left.(Total left 3)

In E2, if the first ball drawn is white, there is 1 white ball left. (Total left 3)

In E3, if the first ball drawn is white, there are 2 white balls left. (Total left 3.)

In E4, if the first ball drawn is white, there are still 3 more white balls left. (Total left 3)

So, the final answer should be [(1/4*0/3) + (1/4*1/3) + (1/4*2/3) + (1/4*3/3)]

Which equals to 1/12 + 2/12 + 3/12 = 1/2

I believe this to be the correct answer however the book has got 4/9 for the answer. Where exactly did I go wrong?

(Thanks for your time and sorry if there are any formatting errors in question.)

  • $\begingroup$ $E_0, E_1...$ etc are not equi-probable ! $\endgroup$ Feb 13, 2017 at 13:30

1 Answer 1


Hint: We can model this by arranging the ten balls in a line, so that the first four are those moved into the second urn then removed in that order.   Given that the first ball is white, what is the probability that the second ball is too?

That is the short method.

The long method is:

Let $E_x$ be the event of $x$ white balls among the four balls drawn from five white and five black, for $x\in\{0,1,2,3,4\}$.   This corresponds to a hypergeometric distribution.

$$\mathsf P(E_x) ~=~ \dfrac{\dbinom 5 x\dbinom 5 {4-x}}{\dbinom {10}4}\cdot\mathbf 1_{x\in\{0,1,2,3,4\}}$$

The events $E_0,E_1,E_2,E_3,E_4$ are thus not equally probable.

Let $A, B$ be the event of drawing white balls in two consecutive draws from among those four.

$$\mathsf P(B\mid A)=\dfrac{\sum_{x=2}^4 \mathsf P(E_x)~\mathsf P(A\cap B\mid E_x)}{\sum_{x=1}^4 \mathsf P(E_x)~\mathsf P(A\mid E_x)}=\dfrac{\sum_{x=2}^4 \binom 5x\binom 5{4-x}\binom x 2/\binom 4 2}{\sum_{x=1}^4 \binom 5x\binom 5{4-x}\binom x 1/\binom 4 1}$$


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