In a large population, 60% of the people have been vaccinated. If 4 people are randomly selected, what is the probability that AT LEAST ONE of them has been vaccinated?

Give your answer as a decimal to 4 places.

Answer:

0.9744 probability that AT LEAST ONE of them has been vaccinated

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they have been vaccinated, or they have not. The probability of a person having been vaccinated 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.

60% of the people have been vaccinated.

This means that [tex]p = 0.6[/tex]

If 4 people are randomly selected, what is the probability that AT LEAST ONE of them has been vaccinated?

This is

[tex]P(X \geq 1)[/tex] when [tex]n = 4[/tex].

We have that

[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_{4,0}.(0.6)^{0}.(0.4)^{4} = 0.0256[/tex]

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0256 = 0.9744[/tex]

0.9744 probability that AT LEAST ONE of them has been vaccinated

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