Suppose 60% of students eat proper nutritious meals daily, and 20% exercise regularly. Also, there is a probability of 25% that a student does both.

Question

Suppose 60% of students eat proper nutritious meals daily, and 20% exercise regularly. Also, there is a probability of 25% that a student does both. Find the probability that student eats proper nutritious meal or exercises but not both.

Answer 1

Ok. So when we are looking for P(A U B), then we want P(A) + P(B), but we don't want both conditions P(A N B), so our formula is P(A) + P(B) - P(A N B), which, A is the event that a student exercises and B is the event in which a student eats a nutritious meal. So P(A)=.20 and P(B)=.60 and P(A N B)= .25.

So we have:

.20+.60-.25=.55

So the probability that a student exercises or eats nutritious meals but not both is .55. I hope that helps.