Respuesta :
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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Infection rate of 2%
This means that [tex]p = 0.02[/tex]
Thirteen random encounters
This means that [tex]n = 13[/tex]
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
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
We want [tex]P(X \geq 1)[/tex]. Then
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{13,0}.(0.02)^{0}.(0.98)^{13} = 0.7690[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.769 = 0.231[/tex]
23.1% probability of meeting at least one person with the flu
