Consider randomly selecting a student at a large university, and let A be the ev

Question

Consider randomly selecting a student at a large university, and let A be the event that the selected student has a Visa card and B be the analogous event for MasterCard. Suppose that  P(A) = 0.7 and P(B) = 0.4.

A. Could it be the case that P(A B) = 0.5? Pick one:

i. Yes, this is possible. Since B is contained in the event A B, it must be the case that P(B) P(A B) and 0.5 > 0.4 does not violate this requirement.

ii. Yes, this is possible. Since A B is contained in the event B, it must be the case that P(B) P(A B) and 0.5 > 0.4 does not violate this requirement.

iii. No, this is not possible. Since B is equal to A B, it must be the case that P(A B) = P(B). However 0.5 > 0.4 violates this requirement.

iiii. No, this is not possible. Since B is contained in the event A B, it must be the case that P(A B) P(B). However 0.5 > 0.4 violates this requirement.

v. No, this is not possible. Since A B is contained in the event B, it must be the case that P(A B) P(B). However 0.5 > 0.4 violates this requirement.

B. From now on, suppose that P(A B) = 0.3. What is the probability that the selected student has at least one of these two types of cards?

C. What is the probability that the selected student has neither type of card?

D. In terms of A and B, the event that the selected student has a Visa card but not a MasterCard is A B' . Calculate the probability of this event.

E. Calculate the probability that the selected student has exactly one of the two types of cards.

Explanation / Answer

A : the selected student has a Visa card

B : the selected student has a MasterCard.

P(A) = 0.7 and P(B) = 0.4

a) P(A and B) can not be 0.5 because P(B)=0.4 and it is less than P(A and B) which is not possible so correct option is v. No, this is not possible. Since A B is contained in the event B, it must be the case that P(A B) P(B). However 0.5 > 0.4 violates this requirement.

b) Given that P(A and B) = 0.3 , P(A or B)=P(A)+P(B)-P(A and B)=0.7+0.4-0.3=0.8

c) P(neither)=1-P(A or B)=1-0.8= 0.2

d) P(A and not B)=P(A or B)-P(B)=0.8-0.4=0.4

e) P(exactly one of 2 types)= P(A or B)-2*P(A and B)=0.8-2*0.3=0.2 ( because we need to subtract P(A and B) twice as it is common between the two)

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.