Question

The probability that a person will develop the flu after getting a flu shot is 0.04. In a random sample of 100 people in a community who got a flu shot, what is the probability that 5 or more of the 100 people will get the flu

Answer 1

Answer:

0.3711 = 37.11% probability that 5 or more of the 100 people will get the flu

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they will develop the flu after getting the shot, or they will not. The probability of a person developing the flu after getting the shot is independent of any other person, which means that the binomial probability distribution is used 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.

The probability that a person will develop the flu after getting a flu shot is 0.04.

This means that $p = 0.04$

Random sample of 100 people:

This means that $n = 100$

What is the probability that 5 or more of the 100 people will get the flu?

This is:

$P(X \geq 5) = 1 - P(X < 5)$

In which

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

So

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

$P(X = 0) = C_{100,0}.(0.04)^{0}.(0.96)^{100} = 0.0169$

$P(X = 1) = C_{100,1}.(0.04)^{1}.(0.96)^{99} = 0.0703$

$P(X = 2) = C_{100,2}.(0.04)^{2}.(0.96)^{98} = 0.1450$

$P(X = 3) = C_{100,3}.(0.04)^{3}.(0.96)^{97} = 0.1973$

$P(X = 4) = C_{100,4}.(0.04)^{4}.(0.96)^{96} = 0.1994$

Then

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0169 + 0.0703 + 0.1450 + 0.1973 + 0.1994 = 0.6289$

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.6289 = 0.3711$

0.3711 = 37.11% probability that 5 or more of the 100 people will get the flu