Question

A graduate class consists of six students. What is the probability that at least 5 of them are born either in April or in October? (Assume, for simplicity, that every year has 365 days.)

Answer 1

Answer:

0.0671% probability that at least 5 of them are born either in April or in October

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they were born in April or October, or they were not. The probabilty of a student being born in April or October is independent of other students. 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.

Probability of a student being born in April or October

April has 30 days, October 31

The year has 365 days. So

$p = \frac{61}{365} = 0.167$

A graduate class consists of six students.

This means that $n = 6$

What is the probability that at least 5 of them are born either in April or in October?

$P(X \geq 5) = P(X = 5) + P(X = 6)$

In which

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

$P(X = 5) = C_{6,5}.(0.167)^{5}.(0.833)^{1} = 0.000649$

$P(X = 6) = C_{6,6}.(0.167)^{6}.(0.833)^{0} = 0.000022$

$P(X \geq 5) = P(X = 5) + P(X = 6) = 0.000649 + 0.000022 = 0.000671$

0.0671% probability that at least 5 of them are born either in April or in October