Q.4.10 - A First Course in Probability

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 $P\{X\le k\}$.

Step-by-Step Solution

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 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$