andrealeon2000
14.02.2020 •
Mathematics
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?
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Ответ:
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
Ответ:
idk
step-by-step explanation: