Question

question 18
in a large population, 69% of the people have been vaccinated. if 5 people are randomly selected, what is the probability that at least one of them has been vaccinated?
give your answer as a decimal (to at least 3 places) or fraction.

Explanation:

Step1: Find the probability that none of them is vaccinated

The probability that a person is not vaccinated is \(1 - 0.69=0.31\).
Since the selections are independent, the probability that all 5 people are not vaccinated is \(P(\text{none})=(0.31)^5\).
Using the formula \(a^n\) where \(a = 0.31\) and \(n = 5\), we have \((0.31)^5=0.31\times0.31\times0.31\times0.31\times0.31 = 0.00008875031\).

Step2: Use the complement rule

The probability that at least one is vaccinated is \(P(\text{at least one})=1 - P(\text{none})\).
Substitute \(P(\text{none})=(0.31)^5\) into the formula: \(P(\text{at least one})=1-(0.31)^5\).
\(1 - 0.00008875031=0.99991124969\approx0.99991\) (rounded to 5 decimal - places)

Answer:

\(0.99991\)