Question

there are 15 students performing in the talent show, 11 of whom will play an instrument. if 12 students are randomly chosen to perform during the first section of the show, what is the probability that exactly 9 of the chosen students will play an instrument? write your answer as a decimal rounded to four decimal places.

Explanation:

Step1: Calculate combination values

We use the combination formula $C(n,k)=\frac{n!}{k!(n - k)!}$. The total number of ways to choose 12 students out of 15 is $C(15,12)=\frac{15!}{12!(15 - 12)!}=\frac{15!}{12!3!}=\frac{15\times14\times13}{3\times2\times1}=455$. The number of ways to choose 9 students who play an instrument out of 11 is $C(11,9)=\frac{11!}{9!(11 - 9)!}=\frac{11!}{9!2!}=\frac{11\times10}{2\times1}=55$. The number of ways to choose the remaining $12 - 9 = 3$ students from the $15-11 = 4$ non - instrument - playing students is $C(4,3)=\frac{4!}{3!(4 - 3)!}=\frac{4!}{3!1!}=4$.

Step2: Calculate the probability

The probability $P$ that exactly 9 of the 12 chosen students play an instrument is given by the hypergeometric probability formula $P=\frac{C(11,9)\times C(4,3)}{C(15,12)}$. Substituting the values we found above, we get $P=\frac{55\times4}{455}=\frac{220}{455}\approx0.4835$.

Answer:

$0.4835$