Question

A student is applying to the University of Florida (UF) and Florida State (FSU).

There is a 40% chance of being accepted at FSU. If the student is accepted at FSU, the probability of being accepted at UF is 60%. If the student is not accepted at FSU there is an 90% chance of non-acceptance at UF.

What is the probability that a student is accepted at FSU or is accepted at UF?

Answer 1

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Explanation:

To find the probability that a student is accepted at FSU or accepted at UF, we can use the concept of conditional probability and the law of total probability.

Let's denote the events as follows:

A: Accepted at FSU

B: Accepted at UF

We need to find P(A or B), which can be calculated as the sum of the probabilities of each event minus the probability of their intersection:

P(A or B) = P(A) + P(B) - P(A and B)

Given the information provided, we can calculate the probabilities:

P(A) = 0.4 (40% chance of being accepted at FSU)

P(B|A) = 0.6 (60% chance of being accepted at UF if accepted at FSU)

P(B|A') = 0.9 (90% chance of non-acceptance at UF if not accepted at FSU)

P(A and B) = P(A) * P(B|A) = 0.4 * 0.6 = 0.24 (probability of being accepted at both FSU and UF)

Now we can substitute these values into the formula:

P(A or B) = P(A) + P(B) - P(A and B)

= 0.4 + (1 - 0.4) * P(B|A') - P(A and B)

= 0.4 + 0.6 * 0.9 - 0.24

= 0.4 + 0.54 - 0.24

= 0.7

Therefore, the probability that a student is accepted at FSU or accepted at UF is 0.7, or 70%.