Question

Meeting at least one person with the flu in thirteen random encounters on campus when the infection rate is 2% (2 in 100 people have the flu)

Answer 1

Answer:

23.1% probability of meeting at least one person with the flu

Step-by-step explanation:

For each encounter, there are only two possible outcomes. Either the person has the flu, or the person does not. The probability of a person having the flu is independent of any other person. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

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

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

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

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

And p is the probability of X happening.

Infection rate of 2%

This means that $p = 0.02$

Thirteen random encounters

This means that $n = 13$

Probability of meeting at least one person with the flu

Either you meet none, or you meet at least one. The sum of the probabilities of these outcomes is 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$. Then

$P(X \geq 1) = 1 - P(X = 0)$

In which

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

$P(X = 0) = C_{13,0}.(0.02)^{0}.(0.98)^{13} = 0.7690$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.769 = 0.231$

23.1% probability of meeting at least one person with the flu