Question

It is known that 60% of the students at a large university have a job and 40% do not have a job. If three of these students are randomly selected what is the probability all three have jobs?
a. 0.360
b. 0.216
c. 0.648
d. 0.060

Answer 1

Answer:

b. 0.216

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they have a job, or they do not. The probability of a student having a job is independent of other students. 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.

60% of the students at a large university have a job

This means that $p = 0.6$

If three of these students are randomly selected what is the probability all three have jobs?

This is $P(X = 3)$ when $n = 3$. So

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

$P(X = 3) = C_{3,3}.(0.6)^{3}.(0.4)^{0} = 0.216$

So the correct answer is:

b. 0.216