Return to the credit card scenario of Exercise 12 (Section 2.2), and let C be th

Question

Return to the credit card scenario of Exercise 12 (Section 2.2), and let C be the event that the selected student has an American Express card. In addition to P(A) = .6, P(B) = .4, and P(A n B) = .3, suppose that P(C) = .2, P(A n C) = .15, P(B n C) = .1, and P(A n B n C) = .08.

A. What is the probability that the selected student has at least one of the three types of cards?

B. What is the probability that the selected student has both a Visa card and a MasterCard but not an American Express card?

C. Calculate and interpret P(B | A) and also P(A | B).

D. If we learn that the selected student has an American Express card, what is the probability that she or he also has both a Visa card and a MasterCard?

E. Given that the selected student has an American Express card, what is the probability that she or he has at least one of the other two types of cards?

Explanation / Answer

a)probability that the selected student has at least one of the three types of cards

P(AUBUC) = P(A) +P(B)+P(C)-P(AnB)-P(AnC)-P(BnC)+P(AnBbC)

=0.6+0.4+0.2-0.3-0.15-0.1+0.08=0.73

b)probability that the selected student has both a Visa card and a MasterCard but not an American Express card

=P(AnB)-P(AnBnC) =0.3-0.08 =0.22

c)P(B|A) =P(A n B)/P(A)=0.3/0.6=0.5 ( probability of having a Master card given have Visa Card)

P(A|B)=P(AnB)/P(B) =0.3/0.4=0.75 ( probability of having a Visa card given have master card)

d)P(AnB|C)= P(AnBnC)/P(C)= 0.08/0.2 =8/20=2/5 =0.4

e) P(AUB|C) =P((AUB)nC)/P(C) =(0.15+0.1-0.08)/0.2 =0.17/0.20=17/20=0.85

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