Question

There are 12 people waiting for extra tickets to a comedy show. Three tickets become available and are randomly assigned. How many different groupings of people can be given the tickets? What is the probability you and your two friends are the 3 people that receive the tickets?

Answer 1

Let's designate an event $\Alpha=$ { among the ticket holders will be me and 2 of my friends }.

Among 12 people in a group of 3 ticket holders can choose

$C$$\begin{matrix} 3 \\ 12 \end{matrix}$ = $\frac{12!}{3!(12-3)!}$ = $\frac{12!}{3! 9!}$ = $\frac{10*11*12}{3!}$ = $\frac{10*11*2}{1}$ = 220 ways.

3 ticket holders out of 3 friends (my company) can be chosen

$C$$\begin{matrix} 3 \\ 3 \end{matrix}$ = $\frac{3!}{3!(3-3)!}$ = $\frac{3!}{3!}$ = 1 way, and 0 "strangers" will receive tickets

$C$$\begin{matrix} 0 \\ 9 \end{matrix}$ = $\frac{9!}{0!(9-0)!}$ = $\frac{9!}{9!}$ = 1.

According to the formula

$P(A) = \frac{C\begin{matrix} 3 \\ 3 \end{matrix} * C\begin{matrix} 0 \\ 9 \end{matrix}}{C\begin{matrix} 3 \\ 12 \end{matrix}} = \frac{1}{220}$ $\approx 0,0045$ .

Answer: 220 , $\frac{1}{220}.$