Q. 2.15 - A First Course in Probability

Question

An urn contains $M$white and $N$black balls. If a random sample of size $r$is chosen, what is the probability that it contains exactly $k$white balls?

Step-by-Step Solution

Answer

Define the outcome space of equally probable combinations.

$P\left(A\right)=\frac{\left(\begin{array}{c}N\\ k\end{array}\right)\left(\begin{array}{c}M\\ r-k\end{array}\right)}{\left(\begin{array}{c}N+M\\ r\end{array}\right)}$

Step 1 Given information.

An urn contains $M$white and $N$black balls. If a random sample size$r$ is chosen.

Step 2 Explanation.

Choose randomly $r$balls from $N$black and $M$white balls. What is the probability that precisely $k$white balls are chosen?

We define the sample space$S$.

$S$- contains every possible choice of$r$ $M+N$distinct objects.

Every element $S$is equally probable. So from Axioms, the probability of an event$A\subseteq S$.

$P\left(A\right)=\frac{\left|A\right|}{\left|S\right|}$

Using combinations from the chapter$1,$$\left|S\right|=\left(\begin{array}{c}M+N\\ r\end{array}\right)$.

Say that$A$ is an event where precisely$k$ white balls are chosen.

The number of combinations of $k$white balls $N$is$\left(\begin{array}{l}N\\ k\end{array}\right)$.

The number of combinations for choosing the remaining $r-k$black balls from $M$is$\left(\begin{array}{c}M\\ c-k\end{array}\right)$.

And since we do not differentiate different permutations in choosing, the final result would be:

$\left|A\right|=\left(\begin{array}{c}N\\ k\end{array}\right)\left(\begin{array}{c}M\\ r-k\end{array}\right)$

So using the formula for the probability$\left(1\right):$

$P\left(A\right)=\frac{\left(\begin{array}{c}N\\ k\end{array}\right)\left(\begin{array}{c}M\\ r-k\end{array}\right)}{\left(\begin{array}{c}N+M\\ r\end{array}\right)}$

This result can't be much simplified.