Question

A school newspaper reporter decides to randomly survey 18 students to see if they will attend Tet festivities this year. Based on past years, she knows that 11% of students attend Tet festivities. We are interested in the number of students who will attend the festivities. Find the probability that more than 2 students will attend.

Answer 1

Answer:

There is a 13.80% probability that more than 2 students will attend.

Step-by-step explanation:

For each student surveyed, there are only two possible outcomes. Either they will attend the Tet festivities this year, or they will not. This means that we can solve this problem using binomial probability distribution concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

11% of students attend Tet festivities, so $p = 0.11$.

18 students are surveyed, so $n = 18$.

Find the probability that more than 2 students will attend.

This is $P(X > 2)$, that is:

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{18,0}.(0.11)^{0}.(0.89)^{16} = 0.1550$

$P(X = 1) = C_{18,1}.(0.11)^{1}.(0.89)^{15} = 0.3448$

$P(X = 2) = C_{18,2}.(0.11)^{2}.(0.89)^{14} = 0.3622$

So

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1550 + 0.3448 + 0.3622 = 0.8620$

Finally

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.8620 = 0.1380$

There is a 13.80% probability that more than 2 students will attend.