Question

Assume that there are an equal number of births in each month so that the probability is that a person chosen at random was born in May. A group of 20 friends meet monthly and celebrate the birthdays for that month. What is the probability that at the May celebration, exactly two members of the group have May birthdays

Answer 1

Answer:

0.2773 = 27.73% probability that at the May celebration, exactly two members of the group have May birthdays

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they have a birthday in May, or they do not. The probability of a person having a birthday in May is independent of any other person. This 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.

Probability of a person being in May:

May has 31 days in a year of 365. So

$p = \frac{31}{365} = 0.0849$

Group of 20 friends:

This means that $n = 20$

What is the probability that at the May celebration, exactly two members of the group have May birthdays?

This is P(X = 2).

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

$P(X = 2) = C_{20,2}.(0.0849)^{2}.(0.9151)^{18} = 0.2773$

0.2773 = 27.73% probability that at the May celebration, exactly two members of the group have May birthdays