Question

Consider randomly selecting a student at a large university. Let A be the event that the selected student has a Visa card, let B be the analogous event for MasterCard, and let C be the event that the selected student has an American Express card. Suppose that
P(A) = 0.6, P(B) = 0.4, and P(A ? B) = 0.3, suppose that P(C) = 0.2, P(A ? C) = 0.11, P(B ? C) = 0.1, and P(A ? B ? C) = 0.07.

(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 P(B | A) and P(A | B).

P(B | A) =

P(A | B) =

(d) Interpret P(B | A) and P(A | B). (Select all that apply.)

P(A | B) is the probability that a student does not have a MasterCard or a Visa card.

P(B | A) is the probability that given that a student has a Visa card, they also have a MasterCard.

P(A | B) is the probability that given that a student has a Visa card, they also have a MasterCard.

P(B | A) is the probability that given that a student has a MasterCard, they also have a Visa card.

P(B | A) is the probability that a student does not have a MasterCard or a Visa card.

P(A | B) is the probability that given that a student has a MasterCard, they also have a Visa card.

(e) 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?

(f) 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?

Answer 1

Answer:

a. 0.76

b. 0.23

c. 0.5

d. p(B/A) is the probability that given that a student has a visa card, they also have a master card

p(A/B) is the probability that given a student has a master card, they also have a visa card

e. 0.35

f. 0.31

Step-by-step explanation:

a. p(AUBUC)= P(A)+P(B)+P(C)-P(AnB)-P(AnC)-P(BnC)+P(AnBnC)

                    =0.6+0.4+0.2-0.3-0.11-0.1+0.07= 0.76

b. P(AnBnC')= P(AnB)-P(AnBnC)

                    =0.3-0.07= 0.23

c. P(B/A)= P(AnB)/P(A)

             =0.3/O.6= 0.5

e. P((AnB)/C))= P((AnB)nC)/P(C)

                     =P(AnBnC)/P(C)

                     =0.07/0.2= 0.35

f. P((AUB)/C)= P((AUB)nC)/P(C)

                    =(P(AnC) U P(BnC))/P(C)

                     =(0.11+0.1)/0.2

                     =0.21/0.2 = 0.31