# Probability

#### laprec

##### New member
Kindly assist with these questions:
Data showed that 22% of people in a small town was infected with the COVID-19 virus. A random sample of six residents from this town was selected.
1) What is the probability that exactly two of these residents was infected?
2) What is the probability that at most 1 of these residents was infected?
Thank you

#### Country Boy

##### Well-known member
MHB Math Helper
If the probability a given person has the virus is p then the probability that person does not have it is 1- p. First imagine putting the six people in a given order. The probability the first person has the virus is p, the probability the next person has the virus is p, the probability the third person does NOT have the virus is 1- p, the probability the fourth person does not is1-p, and the probability the fifth and sixth persons do not is 1-p.

The probability the first two people have the virus and the other four do not, in that order, is $p^2(1-p)^4$.

There are $\frac{6!}{2!4!}= \frac{6(5)}{2}= 15$ different orders of those two people who have the virus and four who do not so the probability two of a random six people have the diease and four do not is $15p^2(1- p)^4$.

In your problem, of course, p= 0.22.

"At most one" means "either one or none". By the same argument as above, the probability none of the people has the virus is $(1- p)^6$ and the probability exactly one has it is $6p(1-p)^5$. Since "none" and "exactly one" are "mutually exclusive", the probability of "at most one" is the sum of those.