Question

Balls are randomly withdrawn, one at a time without replacement, from an urn that initially has N white and M black balls. Find the probability that n white balls are drawn before m black balls,$n\le N,m\le M$

Short answer

In the given information the required probability is$P(X\ge n)=$$\sum _{k=n}^{n+m-1}\frac{\left(\begin{array}{l}N\\ k\end{array}\right)\left(\begin{array}{c}M\\ n+m-1-k\end{array}\right)}{\left(\begin{array}{c}N+M\\ n+m-1\end{array}\right)}$

Step-by-step solution

Given information

Consider this idea. There will be drawn $n$white balls before $m$black balls if and only if there are at least $n$white balls within $n+m-1$drawn balls. This is because the fact that in that case, there is $m$or less black balls within $n+m-1$drawn balls, so we have satisfied our condition. If we mark with X the number of white balls drawn within$n+m-1$drawn balls, we have that X has Hypergeometric distribution.

Step 2:Calculation

$P(X\ge n)=\sum _{k=n}^{n+m-1}P(X=n)$

= $\sum _{k=n}^{n+m-1}\frac{\left(\begin{array}{l}N\\ k\end{array}\right)\left(\begin{array}{c}M\\ n+m-1-k\end{array}\right)}{\left(\begin{array}{c}N+M\\ n+m-1\end{array}\right)}$

Final answer

The required probability is$P\left(X\ge n\right)=$$\sum _{k=n}^{n+m-1}\frac{\left(\begin{array}{c}N\\ k\end{array}\right)\left(\begin{array}{c}M\\ n+m-1-k\end{array}\right)}{\left(\begin{array}{c}N+M\\ n+m-1\end{array}\right)}$