Similar Question
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) = 0.5; P(B) = 0.4; and P(A N B) = 0.25. (A N B) = The intersection of A and B.
a) If the randomly selected student has a Visa card, what is the probability they have a
MasterCard?
b) Calculate and interpret P(B'|A).
c) Calculate and interpret P(A|B)
d) If the randomly selected student has a MasterCard, what is the probability they do not
have a Visa card?
e) Given that the selected individual has at least one card, what is the probability that he
or she has a Visa card?
Answer:
a)
P(MasterCard|Visa) = P(Visa and MasterCard)/P(Visa) = P(B|A) =
P(A n B)/P(A) =
.25/.5 =
1/2 =
.5
b) P(B'|A) = P(no Master Card|Visa) or the probability of no MasterCard given a Visa = P(no Master Card and Visa)/P(Visa) =
P(B' n A)/P(A)
P((B'UB) n A) = P(A)
Therefore, P(B'A) + P(BA) = P(A)
P(B'A) = P(A) - P(BA) = .5 - .25 = .25
Therefore, P(B'|A) = P(B' n A)/P(A) = .25/.5 = 1/2 = .5
c) P(A|B) = P(Visa|MasterCard) =the probability of a Visa given a MasterCard =
P(Visa and MasterCard)/P(MasterCard) = .25/.4 = 5/5 = .625
d) P(no Visa|MasterCard) = P(A'|B) = P(A'B)/P(B)
P(A'B) + P(AB) = P(B)
Therefore, P(A'B) = P(B) - P(AB) = .4 - .25 = .15
P(A'B)/P(B) = .15/.4 = 3/8 = .375
e) P(Visa Card|Visa Card or MasterCard) = P(Visa Card and (Visa Card or MasterCard))/P(Visa Card or MasterCard) =
P(Visa Card)/P(Visa Card or MasterCard)
(if A is a subset of B, A n B = A, and clearly having a Visa Card is a subset of having a Visa Card or MasterCard)
P(Visa Card or MasterCard) = P(Visa Card) + P(MasterCard) - P(Visa Card or MasterCard) = .4 + .5 - .25 = .65
Then, P(Visa Card)/P(Visa Card or MasterCard) =
.5/.65 = 10/13 ≈ 0.769230769230769
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