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One bag contains $n_1$ white and $n_2$ red balls. Another bag contains $m_1$ white and $m_2$ red balls. $N$ balls are drawn at random from the first bag and transferred to the second. Then $M$ balls are drawn at random from second bag.

What is the probability that

  • exactly $x$ balls are white?

  • $x$ balls or more are white?

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Homework? What approaches have you tried? –  cardinal May 10 '11 at 23:09
    
@cardinal: See here concerning your first point. The second point stands, of course. –  t.b. May 11 '11 at 1:12
    
@Theo, thanks. Now I remember that thread. –  cardinal May 11 '11 at 3:21

1 Answer 1

up vote 2 down vote accepted

Let $n=n_1+n_2$ denote the total number of balls in the first bag and $m=m_1+m_2$ the total number of balls in the second bag. Let $N_1$ denote the number of white balls transferred to the second bag and $N_2$ the number of red balls transferred to the second bag. Let $M_1$ denote the number of white balls drawn from the second bag and $M_2$ the number of red balls drawn from the second bag. Hence $N_1$, $N_2$, $M_1$ and $M_2$ are random but $N=N_1+N_2$ and $M=M_1+M_2$ are deterministic.

Conditionally on $(N_1,N_2)$, $M_1$ is the number of white balls in a subset of $M$ balls chosen from $m_1+N_1$ white balls and $m_2+N_2$ red balls, hence $$ P(M_1=x|N_1=y)={m_1+y\choose x}{m_2+N-y\choose M-x}{m+N\choose M}^{-1}. $$ Likewise, $$ P(N_1=y)={n_1 \choose y}{n_2\choose N-y}{n\choose N}^{-1}. $$ Hence $$ P(M_1=x)=\sum_y{n_1\choose y}{n_2\choose N-y}{m_1+y\choose x}{m_2+N-y\choose M-x}{n\choose N}^{-1}{m+N\choose M}^{-1}. $$ The expectation is simpler, since one can get directly $$ E(M_1)=\frac{M}{n}\frac{nm_1+n_1N}{m+N}. $$

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Thank you Didier for your answer. I must miss something: I do not understand how P(N1=y) and P(M1=x) do NOT depend on m (the total number of balls in the second bag). Where am I wrong? –  Jean-Pierre May 15 '11 at 15:20
    
Thank you. It is ok now. –  Jean-Pierre May 21 '11 at 1:06

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