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Varvara68
2 months ago
15

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?
Mathematics
1 answer:
Zina [12.3K]2 months ago
7 0

Response:

a. 0.76

b. 0.23

c. 0.5

d. p(B/A) signifies the likelihood that a student with a visa card also possesses a MasterCard.

p(A/B) indicates the probability that a student with a MasterCard also has a visa card.

e. 0.35

f. 0.31

Detailed 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)

        = 0.3/0.6 = 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

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