Question

About 19​% of the population of a large country is hopelessly romantic. If two people are randomly​ selected, what is the probability both are hopelessly romantic​? What is the probability at least one is hopelessly romantic​?

Answer 1

Answer:

3.61% probability that both are hopelessly romantic​.

34.39% probability at least one is hopelessly romantic

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they are hopelessly romantic, or they are not. The probability of a person being hopelessly romantic is independent from other people. 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.

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

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.

19​% of the population of a large country is hopelessly romantic.

This means that $p = 0.19$

Two people are randomly​ selected

This means that $n = 2$

What is the probability both are hopelessly romantic​?

This is P(X = 2). So

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

$P(X = 2) = C_{2,2}.(0.19)^{2}.(0.81)^{0} = 0.0361$

3.61% probability that both are hopelessly romantic​.

What is the probability at least one is hopelessly romantic​?

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

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

$P(X = 1) = C_{2,1}.(0.19)^{1}.(0.81)^{1} = 0.3078$

$P(X = 2) = C_{2,2}.(0.19)^{2}.(0.81)^{0} = 0.0361$

$P(X \geq 1) = P(X = 1) + P(X = 2) = 0.3078 + 0.0361 = 0.3439$

34.39% probability at least one is hopelessly romantic