Question

An urn contains $4$white and $4$ black balls. We randomly choose $4$ balls. If $2$ of them are white and $2$are black, we stop. If not, we replace the balls in the urn and again randomly select$4$ balls. This continues until exactly $2$ of the $4$chosen are white. What is the probability that we shall make exactly $n$ selections

Short answer

$P(Y=n)={\left(\frac{17}{35}\right)}^{n-1}\left(\frac{18}{35}\right)$

Step-by-step solution

Step 1:Given information

An urn contains $4$white and$4$black balls. We randomly choose$4$ balls. If$2$ of them are white and $2$ are black, we stop. If not, we replace the balls in the urn and again randomly select$4$ balls. This continues until exactly $2$of the$4$ chosen are white

Step 2:Hypergeometric probability

Probability of selecting exactly two white balls

$N=$Population size $=4+4=8$

$n=$Number of draws $=4$

$m=$Number of observed successes$=4$

The number of successes among random draws from a finite population with two possible outcomes follows a hypergeometric distribution.

Formula hypergeometric probability:

$P(X=i)=\frac{\left(\begin{array}{c}m\\ i\end{array}\right)\left(\begin{array}{c}N-m\\ n-i\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$

Evaluate the definition of hypergeometric probability at $i=2$(as we are interested in the probability of exactly $2$white balls).

$P(X=2)=\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)\left(\begin{array}{l}8-4\\ 4-2\end{array}\right)}{\left(\begin{array}{l}8\\ 4\end{array}\right)}$

$=\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)\left(\begin{array}{l}4\\ 2\end{array}\right)}{\left(\begin{array}{l}8\\ 4\end{array}\right)}$

$=\frac{6·6}{70}$

$=\frac{36}{70}$

$=\frac{18}{35}$

Step 3:Geometric probability

Probability of $n$selections until exactly two white balls are selected

$p=\text{Probability of success}=P(X=2)=\frac{18}{35}$

The number of trials required until the first success follows a geometric distribution.

Definition geometric probability:

$P(Y=k)=q^{k-1}p=(1-p{)}^{k-1}p$

Evaluate the definition of geometric probability at $k=n$:

$P(Y=n)={\left(1-\frac{18}{35}\right)}^{n-1}\left(\frac{18}{35}\right)={\left(\frac{17}{35}\right)}^{n-1}\left(\frac{18}{35}\right)$

Step 4:Final answer

$P(Y=n)=$${\left(\frac{17}{35}\right)}^{n-1}\left(\frac{18}{35}\right)$