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

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

58.6%

the correct answer is the second option

wth is with the little mtn top and what are you talking about like fix your words and accually make sence with them and i like was in a completly different question like three times like it seems like your ilitteret