Question

About 13​% of the population of a large country is nervous around strangersnervous around strangers. If two people are randomly​ selected, what is the probability both are nervous around strangersnervous around strangers​? What is the probability at least one is nervous around strangersnervous around strangers?

Answer 1

Answer:

1.69% probability both are nervous around strangers.

24.31% probability at least one is nervous around strangers.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they are nervous around strangers, or they are not. The people are chosen randomly, which means that the probability of a person being nervous around strangers is independent from other people. So we use the binomial probability distribution to solve this problem.

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.

About 13​% of the population of a large country is nervous around strangers.

This means that $p = 0.13$.

Two people are randomly​ selected

This means that $n = 2$.

What is the probability both are nervous around strangers?

This is $P(X = 2)$

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

$P(X = 2) = C_{2,2}.(0.13)^{2}.(0.87)^{0} = 0.0169$

1.69% probability both are nervous around strangers.

What is the probability at least one is nervous around strangersnervous around strangers?

This is $P(X \geq 1)$

$P(X \geq 1) = P(X = 1) + P(X = 2)$

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

$P(X = 1) = C_{2,1}.(0.13)^{1}.(0.87)^{1} = 0.2262$

$P(X = 2) = C_{2,2}.(0.13)^{2}.(0.87)^{0} = 0.0169$

$P(X \geq 1) = P(X = 1) + P(X = 2) = 0.2262 + 0.0169 = 0.2431$

24.31% probability at least one is nervous around strangers.