Question

Consider randomly selecting a student at a large university, and let "A" be the event that the selected student has a Visa card and "B" be the analogous event for MasterCard. Suppose that P(A)=0.6 and P(B)=0.4.

a) Could it be the case that P(A and B)=0.5? Why or why not?

Answer 1

Let A denote the event that the selected individual has a Visa credit card and B be the event of having Master card. The probability of occurring event A is 0.6 and the probability of occurring event B is 0.4 that is, .

It is known that. So, it is not possible that the intersection case, .

is the probability of the common outcomes from the P(A) and P(B). but it is given that P(B)=0.4 which means cannot be exceeded from 0.4.

understanding in detail

assuming that P(A n B) means "Probability of A and not B" or P(A AND (NOT B))

a) it should be possible to have P(A n B) = 0.5 because included in P(A) = 0.6 could be instances of randomly selecting a student who has BOTH Visa and MC. So in that case

P(A AND (NOT B)) + P(A AND B) = P(A)

OR . . . 0.5 + 0.1 = 0.6

in other words, the likelihood of a random student having BOTH cards is then P(A AND B) = 0.1.

however, if the notation P(A n B) means P(A conjunction B) or P(A AND B) then the answer is different.

P(A AND B) = 0.5 could not happen given your numbers above because that would imply P(B) >= 0.5.

In words: you can't have the situation where P(B) = 0.4 and P(A AND B) = 0.5, because a student possessing a MC occurs in EVERY instance of P(A AND B), but that's contradictory to P)B) = 0.4.

Clear?

In one statement you would be saying that P(B) = 0.4 and in the other P(B) >= 0.5, which is contradictory.