Question

Consider randomly selecting a student at a large university, and let Abe the event that the selected student has a Visa card and Bbe the analogous event for MasterCard. Suppose that P(A)=.6 and P(B)=.4.

a. Could it be the case that P(A\( \cap \)B)=.5? Why or why not? (Hint:See Exercise 24.)

b. From now on, suppose that P(A\( \cap \)B)=.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 Aand 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 the two types of cards.

Short answer

a. It is not possible that\(P\left( {A \cap B} \right)\)

b. The probability that the selected student has at least one of these two types of cards is 0.7

c. The probability that the selected student has neither type of card is0.3

d.The probability that the selected student has a Visa card but not a MasterCard is 0.3

e. The probability that the selected student has exactly one of the two types of cards 0.4

Step-by-step solution

Given information

Event A represents that the selected student has a Visa card.

Event B represents the selected student who has a Master card.

The probability of event A is \(P\left( A \right) = 0.6\).

The probability of event B is \(P\left( B \right) = 0.4\).

Compute the probability

a.

Since \(A \cap B\) is contained in B, this implies that \(P\left( {A \cap B} \right) \le P\left( B \right)\).

This inequality violates as 0.5>0.4.

Therefore, this is not possible.

b.

The probability is \(P\left( {A \cap B} \right) = 0.3\).

The probability that the selected student has at least one of these two types of cards is computed as,

\(\begin{aligned}P\left( {A \cup B} \right) &= P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\\ &= 0.6 + 0.4 - 0.3\\ &= 0.7\end{aligned}\)

Therefore, the probability that the selected student has at least one of these two types of cards is 0.7.

c.

The probability that the selected student has neither type of card is computed as,

\(\begin{aligned}1 - P\left( {A \cup B} \right) &= 1 - 0.7\\ &= 0.3\end{aligned}\)

Therefore, the probability that the selected student has neither type of card is 0.3.

d.

The event that the selected student has a Visa card but not a MasterCard is represented as \(A \cap B'\).

The probability of the above event is computed as,

\(\begin{aligned}P\left( {A \cap B'} \right) &= P\left( A \right) - P\left( {A \cap B} \right)\\ &= 0.6 - 0.3\\ &= 0.3\end{aligned}\)

Therefore, the probability of the provided event is 0.3.

e.

The probability that the selected student has exactly one of the two types of cards is computed as,

\(\begin{aligned}P\left( {A \cap B'} \right) + P\left( {A' \cap B} \right) &= \left( {P\left( A \right) - P\left( {A \cap B} \right)} \right) + \left( {P\left( B \right) - P\left( {A \cap B} \right)} \right)\\ &= \left( {0.6 - 0.3} \right) + \left( {0.4 - 0.3} \right)\\ &= 0.3 + 0.1\\ &= 0.4\end{aligned}\)

Therefore, the probability that the selected student has exactly one of the two types of cards is 0.4.