5 red and 10 black balls in a bowl, with replacement

Problem A bowl contains $5$ red and $10$ black balls. A ball is picked randomly and the colour is noted. After every pick the ball is placed back, and an extra ball of the same color is added to the bowl. Determine the following:

(i) Given that the first $n$ balls are all black, what is the probability (say $\alpha_n$) that the ($n+1$)-st ball is also black? What is $\lim_{n \to \infty} \alpha_n$ ?

(ii) Given that the second until the $(n+1)$-st ball are all black, what is the probability (say $\beta_n$) that the first drawn ball is black. What is $\lim_{n \to \infty} \beta_n$?

My Attempt: (i) Let $A_i$ be the event that on the $i$th pick the ball is black. Then $$P(A_1) = \frac{10}{15}, P(A_2) = \frac{11}{16}, P(A_3) = \frac{12}{17}, \ldots$$ Let $B_n$ be the event that the first $n$-balls were all black. Then I computed $$P(B) = P(A_1) P(A_2) \ldots P(A_n)$$ and so $$P(B_{n+1}) = \alpha_n = \big( \frac{10}{15} \big) \big( \frac{11}{16} \big) \ldots \big( \frac{10 + n-1}{15+n-1} \big) \big( \frac{10+n}{15+n} \big).$$ I think that $\lim_{n\to \infty} \alpha_n = 1$, though not sure how to show this.

Is my reasoning correct, and how to solve problem (ii)? Help is appreciated.

• for (i): You are given that the first $n$ are black, so there is no probability involved there. You've got $10+n$ black balls and $5$ red ones so the probability is $a_n=\frac {10+n}{15+n}$. – lulu May 28 '16 at 12:20
• Ah, I see. Thanks. What about (ii)? Saying the first should be black, is same as saying all the balls up to the $(n+1)$st are black. So the probability is $\frac{10+n+1}{15+n+1}$ ? – Kamil May 28 '16 at 12:50
• Second problem is harder. You need to use Bayes' Theorem. Work out the probabilities for both scenarios (start Black, then continue black or start red but then continue black) then see what portion of those are covered by the "start black" scenario. – lulu May 28 '16 at 13:00

$\alpha_n$ is the probability that the $(n+1)$-th draw is black, given that the first $n$ are.   So it is simply $$\alpha_n =\frac{10+n}{15+n}$$

Let $F$ be the event that the first extraction is black. $\quad \mathsf P(F)=\frac{10}{15}\\\quad \mathsf P(F^\complement)=\frac 5{15}\\\therefore \mathsf P(F)=2~\mathsf P(F^\complement)$

Let $S_n$ be the event that the next $n$ extractions are all black. (The second to $(n+1)$-th)

\begin{align}\beta_n = \mathsf P(F \mid S_n) =&~ \dfrac{\mathsf P(S_n\mid F)~\mathsf P(F)}{\mathsf P(S_n\mid F)~\mathsf P(F)+\mathsf P(S_n\mid F^\complement)~\mathsf P(F^\complement)} \\[1ex] =&~ \dfrac{2~\mathsf P(S_n\mid F)}{2~\mathsf P(S_n\mid F)+\mathsf P(S_n\mid F^\complement)} \end{align}

Evaluate the two conditionals and simplify.

• typo: $P(F)=\frac {10}{15}$. – lulu May 28 '16 at 13:01
• What is $F(S_n \mid F^c)$? Do we know that? – Kamil May 28 '16 at 13:07
• @Kamil $\mathsf P(S_n\mid F^\complement)$ is the probability of subsequently drawing $n$ black balls after first drawing (and thus adding) a white ball. \begin{align}\mathsf P(S_n\mid F)~=~\frac{(11+n)!/10!}{(16+n)!/15!}\\[1ex] \mathsf P(S_n\mid F^\complement)~=~\frac{(10+n)!/9!}{(16+n)!/15!}\end{align} – Graham Kemp May 28 '16 at 23:59