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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
60% of the students at a large university have a job
This means that [tex]p = 0.6[/tex]
If three of these students are randomly selected what is the probability all three have jobs?
This is [tex]P(X = 3)[/tex] when [tex]n = 3[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{3,3}.(0.6)^{3}.(0.4)^{0} = 0.216[/tex]
So the correct answer is:
b. 0.216
