Suppose that the probability of getting a job offer (from a specific company) is 0.17. 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 0.07. Given that a person got a campus interview, what is the probability that they get a job offer?

P (X | CI) = 0.7313

Step-by-step explanation:

X = event of getting a job

CI = Campus interview

P (X | CI) = p ( CI ∩ X) / P ( CI ) = P( CI | X)P(X) / P( CI | X)P(X) + P( CI | X')P(X')

P (X | CI) = 0.93 * 0.17 / 0.93 * 0.17 + 0.07 * (1 - 0.17)

P (X | CI) = 0.93 * 0.17 / 0.93 * 0.17 + 0.07 * 0.83

P (X | CI) = 0.1581 / 0.1581 + 0.0581

P (X | CI) = 0.1581 / 0.2162

P (X | CI) = 0.73126735

P (X | CI) = 0.7313

it costs $20 to interview each person for a survey. find the cost to obtain a +- 3% margin of error.

step-by-step explanation:

copy and paste, put it in google

rx and sx

step-by-step explanation: