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 a MasterCard. Suppose that P(A)=.6 and P(B)=.4
1) Could it be the case that P(A∩B)=0.5? Why or why not?

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

3) What is the probability that the selected student has neither type of card?

4) Describe, in terms of A and B, the event that the selected student has a visa card but not a MasterCard, and then calculate the probability of this event? Calculate the probability that the selected student has exactly one of these two types of cards?

Answer 1

Answer:

1) is not possible

2) P(A∪B) = 0.7

3) 1- P(A∪B) =0.3

4) a) C=A∩B' and P(C)= 0.3

b)  P(D)= 0.4

Step-by-step explanation:

1) since the intersection of 2 events cannot be bigger than the smaller event then is not possible that P(A∩B)=0.5 since P(B)=0.4  . Thus the maximum possible value of P(A∩B) is 0.4

2) denoting A= getting Visa card , B= getting MasterCard the probability of getting one of the types of cards is given by

P(A∪B)= P(A)+P(B) - P(A∩B) = 0.6+0.4-0.3 = 0.7

P(A∪B) = 0.7

3) the probability that a student has neither type of card is 1- P(A∪B) = 1-0.7 = 0.3

4) the event C that the selected student has a visa card but not a MasterCard is given by  C=A∩B'  , where B' is the complement of B. Then

P(C)= P(A∩B') = P(A) - P(A∩B) = 0.6 - 0.3 = 0.3

the probability for the event D=a student has exactly one of the cards is

P(D)= P(A∩B') + P(A'∩B) = P(A∪B) - P(A∩B) = 0.7 - 0.3 = 0.4