Respuesta :
Answer:
78.4% probability at least one does not have a job
Step-by-step explanation:
For each student, there are only two possible outcomes. Either they have a job, or they do not have a job. The probability of a student having a job 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.
[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]
Three of these students are randomly selected
This means that [tex]n = 3[/tex]
What is the probability at least one does not have a job?
Either all of them have a job, or at least one does not. The sum of the probabilities of these events is decimal 1. So
[tex]P(X < 3) + P(X = 3) = 1[/tex]
We want P(X < 3). Then
[tex]P(X < 3) = 1 - P(X = 3)[/tex]
In which
[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]
[tex]P(X < 3) = 1 - P(X = 3) = 1 - 0.216 = 0.784[/tex]
78.4% probability at least one does not have a job
