Probability help needed plz ?

Question

an urn contains 2 white balls and 1 black ball. 1 ball at a time is withdrawn, with each ball in urn equally likely to be chosen. when black ball is withdrawn it is automatically returned to the urn, but when white ball ball is withdrawn it is not returned. find the expected no of withdrawals need to empty urn of white balls ?
and how to u work it out. may i add that this is a degree maths question on probability

Answer 1

the key point in this question is that white balls are NOT returned while the black ones are.
So in order to have all the white ones emptied out, we have the following sequences of drawing balls:
<1> WW
<2> WBW
<3> BWW

Now, all u need to do is find the probability of each sequence occuring and add up the probabilities of all the sequences.

e.g. In order to find probability of <2> WBW :
we have (2/3) * (1/2) * (1/2) = (1/6)

How I got the probabilities above:
2/3 --is the probability of drawing a white on first attempt
1/2 --is the probability of drawing a black one on attempt 2 (remember that the white ball is NOT replaced, so we have only two balls remaining)
1/2 --is the probability of drawing a white out of the remaining two balls (note that the black drawn on the previous attempt was replaced, so we still have two balls, one of which is black and the other one ..?)

After u have gotten the probabilities of all 1,2 and 3, add them up to get your answer.

Hope that was clear enough.

EDIT: Just realized, that the question is looking for the number of withdrawals needed. I'm thinking of a way to find that now. Sorry for the misanswering.

Answer 2

Probability to empty urn (PEU)in 2 withdrawals =

= 2/3*1/2 = 1/3

The drawing of a black ball appears in parentheses.

PEU (3) = (1/3)*2/3*1/2 + 2/3*(1/2)*1/2 = 1/9+1/6 = 5/18

PEU (4) = (1/3)*(1/3)*2/3*1/2 + (1/3)*2/3*(1/2)*1/2 +

+ 2/3*(1/2)*(1/2)*1/2 = 1/27 + 1/18 + 1/12 = 19/108

Since 1 - 1/3 - 5/18 - 19/108 <1/3 and

1/3 is largest probability encountered, then PEU (2) more often than any other. However....

PEU (3) is close to 1/3 (5/18), leaving the remaining 7/18 to be divided among the remaining infinite PEU. So roughly, on AVERAGE, we can expect PEU = 3