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, role="math" localid="1648906629368" $1\le i\le b+g,$ is a girl?

Short answer

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

Step-by-step solution

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.

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 $g$availablelocalid="1648907482431" $\left(\begin{array}{l}g\\ 1\end{array}\right)$ways. We arrange the remaining localid="1648907491200" $(b+g-1)$people in the $(b+g-1)$positions. Ipar We divide this by the total arrangements$(b+g)!$.

Explanation.

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