The probability of a randomly selected adult in

Ответ:

caleb19moody

24.07.2021

0.0954 = 9.54% probability that the combined sample tests positive for the virus. This probability is higher than 5%, and thus, it is not unlikely for such a combined sample to test positive.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they test positive, or they do not. The probability of a person testing positive 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 of a randomly selected adult in one country being infected with a certain virus is 0.005.

This means that p = 0.005

Samples from 20 people

This means that n = 20

What is the probability that the combined sample tests positive for the virus? Is it unlikely for such a combined sample to test positive?

Probability of at least one positive test, which is:

P(X geq 1) = 1 - P(X = 0)

In which

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

$P(X = 0) = C_{20,0}.(0.005)^{0}.(0.995)^{20} = 0.9046$

Then

P(X geq 1) = 1 - P(X = 0) = 1 - 0.9046 = 0.0954

0.0954 = 9.54% probability that the combined sample tests positive for the virus. This probability is higher than 5%, and thus, it is not unlikely for such a combined sample to test positive.

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