Applicants who wish to be admitted to a certain professional skill at a large university are required to take a screening test that was devised by an educational testing service. From the past results, the testing service has estimated that 70% of all applicants are eligible for admission and that 92% of those who are eligible for admission pass the exam, whereas 12% of those who are ineligible for admission pass the exam. Using this information, find the probability that an applicant for admis- sion who passed the exam was actually eligible

To find the probability that an applicant who passed the exam was actually eligible, we can make use of Bayes' theorem.

Let's define the following events:
A: The applicant is eligible for admission.
B: The applicant passed the exam.

Given information:
P(A) = 70% or 0.7 (probability that an applicant is eligible)
P(B|A) = 92% or 0.92 (probability that an eligible applicant passes the exam)
P(B|A') = 12% or 0.12 (probability that an ineligible applicant passes the exam)

We need to find P(A|B), which represents the probability that an applicant is eligible given they passed the exam.

Using Bayes' theorem, the formula is:
P(A|B) = (P(B|A) * P(A)) / P(B)

P(B) can be calculated using the law of total probability:
P(B) = P(B|A) * P(A) + P(B|A') * P(A')

Now, let's calculate each part:

P(A') = 1 - P(A) = 1 - 0.7 = 0.3 (probability that an applicant is ineligible)

P(B) = P(B|A) * P(A) + P(B|A') * P(A')
= 0.92 * 0.7 + 0.12 * 0.3
= 0.644 + 0.036
= 0.68

Now we can calculate P(A|B):

P(A|B) = (P(B|A) * P(A)) / P(B)
= (0.92 * 0.7) / 0.68
= 0.644 / 0.68
≈ 0.947

Therefore, the probability that an applicant for admission who passed the exam was actually eligible is approximately 0.947 or 94.7%.