# Boxes and colored balls with replacement

Suppose there are $n+1$ boxes numbered from $0$ to $n$. The $i$-th box contains $i$ white balls and $n-i$ black balls.

A box is chosen randomly and a ball is selected from the box, after that the ball is returned to the box. This procedure is repeated $k$ times.

Find the probability of getting a white ball in the first extraction.

I need help (ideas, suggestions, hints) to solve the exercise. Suppose $A_i=\{\text{a white ball is extracted from the$i$-th box}\}.$ If I consider just one extraction instead of $k$, then for each $A_i$, the probability of getting a white ball is $P(A_i)=\dfrac{i}{n}$. I have no idea how to do it for more than one extraction, I mean, to think of the problem for $k$ extractions.

• The probability of getting a white ball in the first extraction involves only the first extraction hence one can forget the subsequent extractions.
– Did
Aug 25 '14 at 1:52
• Also since you using replacement, each extraction is independent of the other previous extractions, but this fact is not needed for this particular question because of what Did explained above Aug 25 '14 at 1:58
• The answer is $\frac 12$, because all together there are the same number of black and white balls, and each ball has the same probability of being chosen. Aug 25 '14 at 2:41

Let $W$ be the event of extracting a white ball. Let $B=i$ be the event of selecting box $i: \; i\in\{0,..n\}$. If the choice is without bias, then we have a uniformly distributed discrete random variable.
\begin{align}{\sf P}(W) &= \sum_{i=0}^n {\sf P}(B=i){\sf P}(W\mid B=i) \\ & = \sum_{i=0}^{n} \frac 1 {n+1}\times \frac{i}{n} \\ & = \frac 1 {n(n+1)}\times \frac{n(n+1)}{2} \\ & = \frac{1}{2} \end{align}
Remark There are an equal number of black and white balls, so the probability of selecting one at random should indeed be: $1/2$
To find probability of getting a white ball from first extraction, first note we can just look at this in terms of find probability of getting white ball from an extraction (doesn't matter that its first). With this in mind, we can use the law of total probability as such (let $W$ denote event that extracted a white ball, and $B_{i}$ denote event that chose $i$-th box) $$P(W)=\sum_{i=0}^{n}P(W|B_{i})P(B_{i})$$ From this we see first that probability of choosing any box is equal thus $P(B_{i})=\frac{1}{n+1}$, from here if we chose $i$-th box then $P(W|B_{i})=\frac{i}{n}$. Thus putting these two together we have $$P(W)=\sum_{i=0}^{n}\left(\frac{i}{n}\right)\left(\frac{1}{n+1}\right)=\frac{1}{n(n+1)}\sum_{i=0}^{n}i=\frac{1}{n(n+1)}\left(\frac{n(n+1)}{2}\right)=\frac{1}{2}$$