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.
Consider randomly selecting a student at a certain university, and let A denote the event that
the selected individual has a Visa credit card and B be the analogous event for a MasterCard.
Suppose that P(A) = 0.6 and P(B) = 0.4
a) Could it be the case that P(A n B) = 0.5? Why or why not?
I'm not sure how to answer it or prove why it would be true or not.
2 answers
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.
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.
1 answer