One of the most effective recruiting techniques to the Virginia Tech Honors program is getting the students to visit the campus before making their decision. According to recent data, 45% of the potential honors students visit the campus before making their decision. If the student visits, they accept the offer to attend Virginia Tech 95% of the time, but the students that do not visit the campus before making their decision will attend 55% of the time. Knowing a student will enter the Virginia Tech Honors program, what is the probability that they visited the campus before making their decision

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DeanR DeanR · Answer 1 · 2020-05-05T15:10:45+02:00

Wow a real Bayes Theorem question.

Let's name some events.

A - honor student accepts offer to attend VT

V - honor student visits VT

45% of the potential honors students visit the campus before making their decision

P(V) = .45

If the student visits, they accept the offer to attend Virginia Tech 95% of the time

P(A | V) = 0.95

but the students that do not visit the campus before making their decision will attend 55% of the time.

P(A | ¬V) = 0.55

Knowing a student will enter the Virginia Tech Honors program, what is the probability that they visited the campus before making their decision

We're asked for P(V | A)

OK, according to Bayes Theorem

P(V | A) = P(A | V) P(V) / ( P(A | V) P(V) + P(A | ¬V) P(¬V) )

Of course P(¬V) = 1 - P(V) = 1 - .45 = .55

I think we have everything to do the calculation.

P(V | A) = (.95 × .45)/ ( .95 × .45 + .55 × .55 ) = 0.5856164

Answer: 58.6%