Question

A village fete has a children’s running race each year, run in heats of up to ten children. For each heat the first three contestants past the finishing line qualify for the final. There are three prizes in the final for 1st, 2nd and 3rd places. One year 29 children enter the race so there are three heats, of ten, ten and nine children. One year 29 children enter the race so there are three heats, of ten, ten and nine children. 1) What is the probability that three randomly chosen competitors win prizes? 2) What is the probability that two randomly chosen competitors win prizes? 3) How many ways are there to select ten competitors for the first heat? 4) Once the competitors have been selected for the first heat, how many different groups of three qualifiers are possible from this heat

Answer 1

Answer:  1) 1/3,654     2) 3/406     3) 72,684,900,288,000      4) 120

Step-by-step explanation:

1)         First       and        Second         and         Third

  $\dfrac{3\ total\ prizes}{29\ total\ people}\times \dfrac{2\ remaining\ prizes}{28\ remaining\ people}\times \dfrac{1\ remaining\ prize}{27\ remaining\ people}=\dfrac{6}{21,924}\\\\\\.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad =\large\boxed{\dfrac{1}{3,654}}$

2)         First       and          Second

   $\dfrac{3\ total\ prizes}{29\ total\ people}\times \dfrac{2\ remaining\ prizes}{28\ remaining\ people}=\dfrac{6}{812}=\large\boxed{\dfrac{13}{406}}$

$3)\quad \dfrac{29!}{(29-10)!}=\large\boxed{72,684,900,288,000}$

$4)\quad _{10}C_3=\dfrac{10!}{3!(10-3)!}=\large\boxed{120}$