Question

a local high school takes pride in the fact that 91% of its students participate in after school activities.
if 2 students are randomly selected, what is the probability that fewer than 2 of the chosen students participate in after school activities?
write your answer as a decimal rounded to the nearest thousandth.

Explanation:

Step1: Calculate the probability of the complementary event

The probability that a student participates in after - school activities \(p = 0.91\), so the probability that a student does not participate \(q=1 - p=1 - 0.91 = 0.09\).
The probability that fewer than 2 of the 2 chosen students participate in after - school activities is equal to \(1\) minus the probability that exactly 2 students participate.
The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times q^{n - k}\), where \(n = 2\), \(k = 2\), \(C(n,k)=\frac{n!}{k!(n - k)!}\), \(n!=n\times(n - 1)\times\cdots\times1\).
For \(n = 2\), \(k = 2\), \(C(2,2)=\frac{2!}{2!(2 - 2)!}=\frac{2!}{2!×0!}=1\) (since \(0!=1\)).
\(P(X = 2)=C(2,2)\times(0.91)^{2}\times(0.09)^{0}\)
\(P(X = 2)=1\times0.91^{2}\times1=0.8281\)

Step2: Calculate the required probability

Let \(P(X\lt2)\) be the probability that fewer than 2 students participate.
\(P(X\lt2)=1 - P(X = 2)\)
\(P(X\lt2)=1-0.8281 = 0.1719\approx0.172\)

Answer:

\(0.172\)