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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!

Michael
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1 Answers1

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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 $n \rightarrow \infty$.

user980341
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  • Can you explain how you get the general case for B2…B_{n+1}? – b.b.89 Jan 16 '24 at 00:52
  • I would have said that you can deduce it from $\mathbb{P}(B_1\cap ... \cap B_{n-1})=\mathbb{P}(B_2\cap ... \cap B_n)$. The derivation of which should be able to be deduced from the fact that $\mathbb{P}(B_1)=\mathbb{P}(B_n)$ so no matter where you start, the initial probability of the first blue ball is always $\frac{b}{r+b}$ (then you go on and multiply the following probabilities $\mathbb{P}(B_{n+1}|B_n)=\frac{b+d}{r+b+d} and so on) – Leoncino Apr 20 '24 at 11:39