Question

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

Answer 1

To find the probability that a student visited the Virginia Tech Honors campus, we can use conditional probability and Bayes' theorem. The probability that a student visited the campus before making their decision is 58.56%.

Step-by-step explanation:

To find the probability that a student visited the campus before making their decision to enter the Virginia Tech Honors program, we need to use conditional probability. Let A be the event that a student visited the campus, and B be the event that a student accepted the offer to attend Virginia Tech.

We are given the following probabilities:

• P(A) = 45% = 0.45 (probability a student visited the campus)
• P(B|A) = 95% = 0.95 (probability a student accepted the offer given that they visited the campus)
• P(B|A') = 55% = 0.55 (probability a student accepted the offer given that they did not visit the campus)

We want to find P(A|B), the probability that a student visited the campus given that they accepted the offer. We can use Bayes' theorem:

P(A|B) = (P(B|A) * P(A)) / P(B)

To calculate P(B), we need to consider both cases:

• P(B) = P(B|A) * P(A) + P(B|A') * P(A')
• P(A') = 1 - P(A) = 1 - 0.45 = 0.55

Substituting the given values, we can calculate the probability:

P(B) = (0.95 * 0.45) + (0.55 * 0.55) = 0.4275 + 0.3025 = 0.73

Now we can calculate P(A|B):

P(A|B) = (0.95 * 0.45) / 0.73 = 0.4275 / 0.73 = 0.5856

Therefore, the probability that a student visited the campus before making their decision to enter the Virginia Tech Honors program is 0.5856, or 58.56%.

Answer 2

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%