Question

A jar contains $n$chips. Suppose that a boy successively draws a chip from the jar, each time replacing the one drawn before drawing another. The process continues until the boy draws a chip that he has previously drawn. Let$X$ denote the number of draws, and compute its probability mass function.

Short answer

$\frac{k}{n^{k+1}}·\prod _{i=0}^{k-1}(n-i)$

Step-by-step solution

Given information

We have that $X\in \mathrm{\mathbb{N}},X\ge 2$. Event $X>k$means that in first $k$draws we have taken all different chips. Hence

$P(X>k)=\frac{n(n-1)\dots (n-k+1)}{n^k}$

Similarly we have that

$P(X>k+1)=\frac{n(n-1)\dots (n-k)}{n^{k+1}}$

Explanation

Hence. we have that

$P(X=k)=P(X>k)-P(X>k+1)$

$=\frac{n(n-1)\dots (n-k+1)}{n^k}-\frac{n(n-1)\dots (n-k)}{n^{k+1}}$

$=\frac{n(n-1)\dots (n-k+1)}{n^k}\left(1-\frac{n-k}{n}\right)$

$=\frac{k}{n}·\frac{n(n-1)\dots (n-k+1)}{n^k}$

Final answer

The probability mass function is

$P(X=k)=\frac{k}{n^{k+1}}·\prod _{i=0}^{k-1}(n-i)$