Q. 2.19 - A First Course in Probability

Question

An urn contains $n$red and $m$blue balls. They are withdrawn one at a time until a total of$r,r ... n$, red balls have been withdrawn. Find the probability that a total of $k$balls

are withdrawn.

Hint: A total of $k$balls will be withdrawn if there are $r - 1$red balls in the first $k - 1$withdrawal and the kth withdrawal is a red ball.

Step-by-Step Solution

Answer

$P\left(A_k\right)=\frac{\left(\begin{array}{c}n\\ r\end{array}\right)\left(\begin{array}{c}m\\ k-r\end{array}\right)·r}{k·\left(\begin{array}{c}m+n\\ k\end{array}\right)}$

Define the outcome space of equally probable events, and count the number of outcomes in each set.

Step 1 Given Information.

An urn contains $n$red and $m$blue balls. They are withdrawn one at a time until a total of$r,r ... n,$ red balls have been withdrawn.

Step 2 Explanation.

Random experiment:

Choose balls one by one from$m$ blue&$n$red.

Until $r,(r\le n)$red balls are chosen.

Calculate the Probability that the experiment ends after $k$balls are chosen.

It is logical to assume that every permutation of the$n+m$ balls will be the order in which the balls are drawn, only in this assumption, every time we draw a ball, every remaining ball is equally likely to be chosen.

$S$- outcome space which contains all permutations of $n+m$balls Probability of the event $A,A\subseteq S$is

$P\left(A\right)=\frac{\left|A\right|}{\left|S\right|} \text{ where }\left|X\right|\text{ is the number of elements in }X$

Name:

$A_k-r$The -th red ball is drawn in $k$the -th draw.

The wanted probability is$P\left(A_k\right)$.

The number of outcomes $A$is counted as follows:

Consider first the number of results for choosing $r$from $n$red balls that are chosen in the first$k$, and that is$\left(\begin{array}{l}n\\ r\end{array}\right)$.

And for every choice of the red balls, choose$k-r$ from $m$blue balls$\left(\begin{array}{c}m\\ k-r\end{array}\right)$.  

Since we have drawn the last red-ball $k$-th, choose a red ball for that draw$k$ ways.

Permute the $k-1$balls before the $k$-th, and $m+n-k$balls after the$k$-th.

By the basic principle of counting, 

$\left|A_k\right|=\left(\begin{array}{c}n\\ r\end{array}\right)\left(\begin{array}{c}m\\ k-r\end{array}\right)·r·(k-1)!·(m+n-k)!$

Applying the formula$\left(1\right):$

$P\left(A_k\right)=\frac{\left(\begin{array}{c}n\\ r\end{array}\right)\left(\begin{array}{c}m\\ k-r\end{array}\right)·r·(k-1)!·(m+n-k)!}{(m+n)!}=\frac{\left(\begin{array}{c}n\\ r\end{array}\right)\left(\begin{array}{c}m\\ k-r\end{array}\right)·r}{k·\left(\begin{array}{c}m+n\\ k\end{array}\right)}$