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) = .6 and P(B) = .
a. Could it be the case that P(A > B) = .5? Why or why not?
b. From now on, suppose that P(A > 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 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 the two types of cards.
The correct answer and explanation is :
To address the given questions, we need to use the probability theory and basic set operations. Let’s break down each part and work through the solution step-by-step.
Given Information:
- P(A) = Probability that a student has a Visa card = 0.6
- P(B) = Probability that a student has a MasterCard (this is unspecified, so we’ll assume this is provided elsewhere in the full problem or is intended to be determined).
- P(A ∩ B) = Probability that a student has both cards. This is not explicitly given, so we’ll need to work with this assumption and the other information provided.
a. Could it be the case that P(A > B) = 0.5? Why or why not?
P(A > B) typically refers to the probability that event A (having a Visa card) occurs, given that event B (having a MasterCard) does not occur, or some comparison between the two events. However, based on the context of probabilities, P(A > B) doesn’t represent a valid probability expression as it is currently written. Probabilities need to be between 0 and 1, and what we usually compare in probability theory is P(A ∩ B) (both events occurring), P(A ∪ B) (either event occurring), or conditional probabilities (e.g., P(A | B)).
Without clear clarification of P(A > B), we can reasonably conclude that this statement as written doesn’t reflect a standard probability expression. Therefore, P(A > B) = 0.5 is likely not a correct or possible interpretation.
b. If P(A > B) = 0.3, what is the probability that the selected student has at least one of these two types of cards?
We need to calculate P(A ∪ B), the probability that the student has at least one of the two cards. This can be found using the formula for the union of two events:
[
P(A \cup B) = P(A) + P(B) – P(A \cap B)
]
We know P(A) = 0.6, and we are told that P(A > B) = 0.3, which typically refers to the conditional probability that a student has a Visa card but not a MasterCard. Therefore, P(A ∩ B’) = 0.3, where B’ is the complement of B (i.e., the student does not have a MasterCard).
Since P(A ∩ B’) is the probability that the student has a Visa but not a MasterCard, we can use:
[
P(A) = P(A \cap B) + P(A \cap B’)
]
[
0.6 = P(A \cap B) + 0.3
]
Thus, P(A ∩ B) = 0.3.
Now we can substitute into the formula for the union:
[
P(A \cup B) = 0.6 + P(B) – 0.3
]
To fully compute P(A ∪ B), we need the value of P(B), which isn’t provided directly in the problem. However, if P(B) = 0.6 (assuming equal probability for Visa and MasterCard holders):
[
P(A \cup B) = 0.6 + 0.6 – 0.3 = 0.9
]
Therefore, the probability that the student has at least one of these two types of cards is 0.9.
c. What is the probability that the selected student has neither type of card?
The probability that the student has neither card is simply the complement of the probability that the student has at least one of the cards:
[
P(\text{neither}) = 1 – P(A \cup B)
]
Using the previously computed value for P(A ∪ B):
[
P(\text{neither}) = 1 – 0.9 = 0.1
]
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.
The event that the student has a Visa card but not a MasterCard is represented by A ∩ B’ (Visa but not MasterCard). We are given that:
[
P(A \cap B’) = 0.3
]
So, the probability that the student has a Visa card but not a MasterCard is 0.3.
e. Calculate the probability that the selected student has exactly one of the two types of cards.
The probability that the student has exactly one of the two cards is the sum of the probabilities of having only a Visa or only a MasterCard. This is:
[
P(\text{exactly one card}) = P(A \cap B’) + P(B \cap A’)
]
We already know P(A ∩ B’) = 0.3, and similarly, P(B ∩ A’) (having only a MasterCard) can be calculated as:
[
P(B \cap A’) = P(B) – P(A \cap B) = 0.6 – 0.3 = 0.3
]
Therefore, the probability of having exactly one card is:
[
P(\text{exactly one card}) = 0.3 + 0.3 = 0.6
]
Final Answers:
- a. P(A > B) = 0.5 is not a valid probability expression.
- b. P(A ∪ B) = 0.9 (assuming P(B) = 0.6).
- c. P(neither) = 0.1.
- d. P(A ∩ B’) = 0.3.
- e. P(exactly one card) = 0.6.