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) = 0.6 and P(B) = 0.4. a. Could it be the case that P(A ∩ B) = 0.5? Why or why not?
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. 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.
e. Calculate the probability that the selected student has exactly one of these two types of cards

A)P(A ∩ B) cannot be equal to 0.5.

B)P(A ⋃ B) = 0.7

C)P[(A ∪ B)'] = 0.3

D)P(A ∩ B') = 0.3

E)probability that the selected student has exactly one of these two types of cards = 0.4

Step-by-step explanation:

A) We want to find out if P(A ∩ B) = 0.5

Now, this value is not possible because the probability of intersection of two events can not be greater than the probability of each individual event.

We are told that P(A) = 0.6 and P(B) = 0.4.

Due to the fact that the probability of B is less than 0.4 which is less than 0.5, It means that the intersection of A and B can't be greater than 0.4.

Thus, P(A ∩ B) cannot be equal to 0.5.

B) Now, P(A ∩ B) = 0.3

The probability that the selected student has at least one of these two types of cards would be P(A ⋃ B)

Now, this is expressed as;

P(A ⋃ B) = P(A) + P(B) − P(A ∩ B)

Thus,

P(A ⋃ B) = 0.6 + 0.4 - 0.3

P(A ⋃ B) = 0.7

C) Probability that the selected student has neither type of card would be expressed as; P[(A ∪ B) ' ]

Thus is further expressed as;

P[(A ∪ B) ' ] = 1 − P(A ∪ B)

P[(A ∪ B)'] = 1 - 0.7

P[(A ∪ B)'] = 0.3

D) We want to describe in terms of A and B, the event that the selected student has a Visa card but not a MasterCard.

This is simply an intersection of event A and compliment of event B. Thus, it implies that we we will remove the events when student has both types of cards from the events of A. This probability is expressed as P(A ∩ B')

Thus gives;

P(A ∩ B') = P(A) - P(A ∩ B)

P(A ∩ B') = 0.6 - 0.3

P(A ∩ B') = 0.3

E) Now, we want to find the probability that the selected student has exactly one of these two types of cards.

This is simply the addition of intersection of event A with compliment of event B and intersection of event B with compliment of event A . The probability is given by:

P(A ∩ B') + P(A' ∩ B)

Expanding this gives;

P(A ∩ B') + P(B) - P(A ∩ B)

Plugging in the relevant values gives;

0.3 + 0.4 - 0.3 = 0.4