# Consider randomly selecting a

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

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

Answer:

Let A denote the event that the selected individual has a Visacredit card and B be the event of having Master card. Theprobability of occurring event A is 0.6 and the probability ofoccurring 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.**

**understandingin detail**

assuming that P(A n B) means “Probability of A and not B” or P(AAND (NOT B))a) it should be possible to have P(A n B) = 0.5 because included inP(A) = 0.6 could be instances of randomly selecting a student whohas BOTH Visa and MC. So in that caseP(A AND (NOT B)) + P(A AND B) = P(A)OR . . . 0.5 + 0.1 = 0.6in other words, the likelihood of a random student having BOTHcards is then P(A AND B) = 0.1.

however, if the notation P(A n B) means P(A conjunction B) orP(A AND B) then the answer is different.P(A AND B) = 0.5 could not happen given your numbers above becausethat would imply P(B) >= 0.5.In words: you can’t have the situation where P(B) = 0.4 and P(A ANDB) = 0.5, because a student possessing a MC occurs in EVERYinstance 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 theother P(B) >= 0.5, which is contradictory.