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An urn contains r red balls and b blue balls. A ball is chosen at random from the urn, its color is noted, and it is returned together with d more balls of the same color. This repeated indefinitely. a) What is the probability that the first ball drawn given that the second ball drawn is blue? b) Find the probability that the first ball is blue given that the n subsequent drawn balls are all blue. Find the limit of this probability as n tends to infinity.

Thank you very much!

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    $\begingroup$ This site works better if you show what you have tried so far $\endgroup$ – Henry Nov 2 '16 at 7:45
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Define the events $\textit{$R_n:=$"$n$-th ball drawn is red"}$ and $\textit{$B_n:=$"$n$-th ball drawn is blue"}$. Then \begin{align*} \mathsf{P}(B_2)&=\mathsf{P}(B_2 \mid B_1) \mathsf{P}(B_1)+\mathsf{P}(B_2 \mid R_1) \mathsf{P}(R_1) \\ &= \frac{b+d}{r+b+d}\frac{b}{r+b}+\frac{b}{r+b+d}\frac{r}{r+b}=\frac{b}{r+b} \end{align*} and in general $\mathsf{P}(B_n)=\frac{b}{r+b}$. Using that $\mathsf{P}(B_n)$ does not depend on $n$ we get $$ \mathsf{P}(B_1 \mid B_2)= \frac{\mathsf{P}(B_2 \mid B_1)\mathsf{P}(B_1)}{\mathsf{P}(B_2)}=\mathsf{P}(B_2 \mid B_1)=\frac{b+d}{r+b+d} $$ and $$ \mathsf{P}(B_1 \mid B_2 \cap \ldots \cap B_{n+1}) = \mathsf{P}(B_{n+1} \mid B_1 \cap \ldots \cap B_n ) = \frac{b+nd}{r+b+nd}, $$ with limit $1$ as $n \rightarrow \infty$.

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