A job fair was held at the Student Union. 25% of the students who attended received job offers. Of all of the students at the job fair, 40%

Question

A job fair was held at the Student Union. 25% of the students who attended received job offers. Of all of the students at the job fair, 40% were from the College of Business. Among these business students, 50% received job offers. Let J be the event that a student is offered a job. Let B be the event that the student is from the College of Business.

Requried:
a. Are events J and B independent? Why or why not?
b. Are events J and B mutually exclusive? Why or why not?
c. Joe, who is not a business student, attended the job fair. What is the probability that he received a job offer?
d. Another student, Samantha, received a job offer. What is the probability that she is a Business student?

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Thiên Di 4 years 2021-08-11T18:15:56+00:00 1 Answers 11 views 0

Answers ( )

  1. thaiduong
    0
    2021-08-11T18:17:41+00:00
    Reply

    Answer:

    A) Both events are not independent.

    B) Both events are not mutually exclusive

    C) 8.33%

    D) 80%

    Step-by-step explanation:

    A) Both events are not independent. This is because, If B occurs it means that it is very likely that J will occur as well.

    B) Both events are not mutually exclusive. This is because it is possible for both events J and B to occur at the same time.

    C) we want to find the probability that Joe who is not a business student will receive the job offer.

    This is;

    P(J|Not B) = P(J & Not B)/P(Not B)

    Now,

    P(J & Not B) = P(J) – (P(B) × P(J | B))

    25% of the students who attended received job offers. Thus; P(J) = 0.25

    40% were from the College of Business. Thus;

    P(B) = 0.4

    Among the business students, 50% received job offers. Thus;

    P(J|B) = 0.5

    Thus;

    P(J & Not B) = 0.25 – (0.4 × 0.5)

    P(J & Not B) = 0.25 – 0.2

    P(J & Not B) = 0.05

    Since P(B) = 0.4

    Then, P(Not B) = 1 – 0.4 = 0.6

    Thus;

    P(J|Not B) = 0.05/0.6

    P(J|Not B) = 0.0833 = 8.33%

    D) This probability is represented by;

    P(B | J) = P(B & J)/P(J)

    P(B & J) = (P(B) × P(J | B)) = (0.4 × 0.5) = 0.2

    P(B | J) = 0.2/0.25

    P(B | J) = 0.8 = 80%

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