Question

(problem 6.13 page 97) Consider a population in which 80% of males and 60% of females are employed. In this population, 55% of individuals are females. If I pick five persons at random from this population,

what is the probability that no more than one of those chosen is not employed?

Answer 1

There is a 50.77% probability that no more than one of those chosen is not employed.

Explanation:

For each person, there are only two possible outcomes. Either they will be employed, or they will not. 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) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

We have these following percentages:

55% of the individuals are female.

So 45% of the individuals are male.

80% of males are employed. So 20% of males are unemployed.

60% of females are employed. So 40% of females are unemployed.

If I pick five persons at random from this population, what is the probability that no more than one of those chosen is not employed?

Using the binomial distribution, p is the probability that a person is unemployed. 40% of the females and 20% of the males are unemployed. The population is 55% females and 45% males. So

`p = 0.4*0.55 + 0.2*0.45 = 0.31`

There are five persons, so
`n = 5`

What is the probability that no more than one of those chosen is not employed?

`P(X \leq 1) = P(X = 0) + P(X = 1)`

In which

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(5,0).(0.31)^(0).(0.69)^(5) = 0.1564`

`P(X = 1) = C_(5,1).(0.31)^(1).(0.69)^(4) = 0.3513`

Then

`P(X \leq 1) = P(X = 0) + P(X = 1) = 0.1564 + 0.3513 = 0.5077`

There is a 50.77% probability that no more than one of those chosen is not employed.