Q. 2.32 - A First Course in Probability

Question

A group of individuals containing $b$boys and $g$girls is lined up in random order; that is, each of the$(b + g)!$ permutations is assumed to be equally likely. What is the probability that the person in the ith position, $1 \le i \le b + g,$ is a girl?

Step-by-Step Solution

Answer

The probability that the person in the ith position,$1\le i \le b + g,$ a girl is$\frac{g}{b+g}$.

Step 1 Given Information.

A group of individuals containing $b$boys and $g$girls is lined up in random order; that is, each of the $(b + g)!$permutations is assumed to be equally likely.

Step 2 Explanation.

$\frac{\left(\begin{array}{l}g\\ 1\end{array}\right)(b+g-1)!}{(b+g)!}=\frac{g}{b+g}$

We choose one girl from the $g$available$\left(\begin{array}{l}g\\ 1\end{array}\right)$ ways. We arrange the remaining $(b+g-1)$people in the $(b+g-1)$positions. Ipar We divide this by the total arrangements$(b+g)!$.

Step 3 Explanation.

the probability that the person in the ith position, $1\le i\le b + g,$ a girl is$\frac{g}{b+g}$.