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Suppose there are 9 white balls and 3 red balls. A,B,C take turns to pick a ball without replacement . The first person to get a red ball wins.

a) What is the probability that A wins in 1st try?

$\Rightarrow \frac{3}{12}=\frac14$

b) What is the probability that B wins in second try?

$\Rightarrow P(\text{B wins})=P(B \text{ picks a red ball}|\text{A picks a white ball})P(\text{A picks a white ball})=\frac3{11}\frac{9}{12}$ After A has picked a white ball, B has to choose from 3 red balls, among 11 balls remaining

The above answer should be $\frac14$

c) What is the probability that B wins in 5th trial?

$ P(\text{B wins in 5th trial})=P(A_1=W,B_1=W,C_1=W,A_2=W,B_2=R)=\frac{9}{12}\frac{8}{11}\frac{7}{10}\frac{6}{9}\frac{3}{8}$

similar question

DRPR
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  • There is a $\frac 14$ chance that $A$ wins on the first try, and a non-zero chance that $A$ wins on a later try, so your answer to $a$ is clearly incorrect. – lulu May 11 '18 at 14:27
  • @lulu updated.. – DRPR May 11 '18 at 14:31
  • Post edit, your answer to $a$ is now correct. As are your answers to $b$ and $c$. – lulu May 11 '18 at 14:32
  • I am not so sure about b,c since https://math.stackexchange.com/questions/2336591/three-white-and-one-red-ball-probability/2336598#2336598 – DRPR May 11 '18 at 14:34
  • That is a different problem. In that one, there is only one winner and it has an equal probability of being in any slot. Here there are multiple winners. for your problem you are looking at very specific paths. – lulu May 11 '18 at 16:04

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