Question

An urn contains$n$balls numbered $1$through $n$. If you withdraw $m$balls randomly in sequence, each time replacing the ball selected previously, find$P\{X=k\},k=$$1,...,m$

where $X$is the maximum of the $m$chosen numbers.

Hint: First find $P\{X\le k\}$.

Short answer

$P(X\le k)={\left(\frac{k}{n}\right)}^m$

$P(X=k)={\left(\frac{k}{n}\right)}^m-{\left(\frac{k-1}{n}\right)}^m$

Step-by-step solution

Step 1: Given information

An urn contains n balls numbered$1$ through $n$. If you withdraw $m$ balls randomly in sequence, each time replacing the ball selected previously.

Step 2: Explanation

For the beginning, we will find $P(X\ge k)$. Observe that in total, there exist$n^m$sequences of drawn balls. If $X\le k$, that means all balls in the sequence that has been drawn must be less or equal to $k$. There exist $k^m$of these sequences. Hence

$P(X\le k)=\frac{k^m}{n^m}={\left(\frac{k}{n}\right)}^m$

Now, we have that

$P(X=k)=P(X\le k)-P(X\le k-1)={\left(\frac{k}{n}\right)}^m-{\left(\frac{k-1}{n}\right)}^m$

Step 3:Final answer

$P(X\le k)=$${\left(\frac{k}{n}\right)}^m$

$P(X=k)$$={\left(\frac{k}{n}\right)}^m-{\left(\frac{k-1}{n}\right)}^m$