(a) Could it be the case that

P(A\\\\cap B)=0.5

? Why or why not? [Hint: For any two sets

A

and

B

if

A

is a subset of

B

then

P(A)

.]\ Yes, this is possible. Since

B

is contained in the event

A\\\\cap B

, it must be the case that

P(B)

and

0.5>0.3

does not violate this requirement.\ No, this is not possible. Since

B

is contained in the event

A\\\\cap B

, it must be the case that

P(A\\\\cap B)

. However

0.5>0.3

violates this requirement.\ No, this is not possible, Since

A\\\\cap B

is contained in the event

B

, it must be the case that

P(A\\\\cap B)

. However

0.5>0.3

vielates this requirement.\ Yes, this is possible. Since

A\\\\cap B

is contained in the event

B

, it must be the case that

P(B)

and

0.5>0.3

does not violate this requirement.\ No, this is not possible. Since

B

is equal to

A\\\\cap B

, it must be the case that

P(A\\\\cap B)=P(B)

. However

0.5>0.3

violates this requirement.\ (b) From now on, suppose that

P(A\\\\cap B)=0,2

. 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) Describe, in terms of

A

and

B

, the event that the selected student has a Visa card but not a MasterCard.\

A^(')\\\\cap B^(')\ A^(')\\\\cap B^(')\ A^(+)\\\\cup B^(')\ A\\\\cup B^(**)

\ Calculate the probability of this event.\ (e) Calculate the probability that the selected student has exactly one of the two types of cards.

(a) Could it be the case that P(AB)=0.5 ? Why or why not? [Hint: For any two sets A and B if A is a subset of B then P(A)P(B).] Yes, this is possible. Since B is contained in the event AB, it must be the case that P(B)P(AB) and 0.5>0.3 does not violate this requirement. No, this is not possible. Since B is contained in the event. AB, it must be the case that P(AB)P(B). However 0.5>0.3 violates this requirement. No, this is not possible. Since AB is contained in the event B, it must be the case that P(AB)P(B). However 0.5>0.3 vielates this requirement. Yes, this is possible. Since AB is contained in the event B, it must be the case that P(B)P(AB) and 0.5>0.3 does not violate this requirement. No, this is not possibie. Since B is equal to AB, it must be the case that P(AB)=P(B). However 0.5>0.3 violates this requirement. (b) From now on, suppose that P(AB)=0.2. 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) Describe, in terms of A and B, the event that the selected student has a Visa card but not a MasterCard. AB+ AB AB AB Calculate the probability of this event. (e) Calculate the probability that the selected student has exactly one of the two types of cards