Question

Suppose that the probability of getting a job offer (from a specific company) is 0.16. Suppose that we observe empirically that given that an applicant for the job was in fact offered a job, the probability of an on campus interview was 0.93. Also, we observe empirically that given the group that were not offered a job the probability of an on campus interview was of 0.08. Given that a person got a campus interview, what is the probability that they get a job offer? (Answer to five decimal places).

Answer 1

Answer:

0.68889

Step-by-step explanation:

This can be formulated as the following question:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

So

Given that a person got a campus interview, what is the probability that they get a job offer?

P(B) is the probability of getting a job offer. So $P(B) = 0.16$

P(A/B) is the probability of getting a campus interview, given that they get a job offer. So P(A/B) = 0.93.

P(A) is the probability ofgetting a campus interview. Of those 16% who got a job, 93% had a campus interview. Of those 84% who did not get a job, 8% had a campus interview. So

$P(A) = 0.16*0.93 + 0.84*0.08 = 0.216$

Finally

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.16*0.93}{0.216} = 0.68889$