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Chapter 2: Problem 14

Consider randomly selecting a student at a certain university, and let \(A\) denote the event that the selected individual has a Visa credit card and \(B\) be the analogous event for a MasterCard. Suppose that \(P(A)=.5, P(B)=.4\), and \(P(A \cap B)=.25\). a. Compute the probability that the selected individual has at least one of the two types of cards (i.e., the probability of the event \(A \cup B\) ). b. What is the probability that the selected individual has neither type of card? c. 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.

Short Answer

Expert verified
a. 0.65; b. 0.35; c. 0.25

Step by step solution

01

Calculate the probability of having at least one type of card

To find the probability of the event that a student has at least one type of credit card, we use the formula for the union of two events: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). Substitute the given probabilities: \( P(A \cup B) = 0.5 + 0.4 - 0.25 = 0.65 \). Therefore, the probability that the student has at least one type of card is 0.65.
02

Calculate the probability of having neither type of card

The probability of having neither type of card is the complement of the probability of having at least one type of card. We can express this as \( P((A \cup B)^c) = 1 - P(A \cup B) \). Using the result from Step 1, we have \( P((A \cup B)^c) = 1 - 0.65 = 0.35 \). Thus, the probability that the student has neither type of card is 0.35.
03

Describe and calculate the probability of having only a Visa card

The event that a student has a Visa card but not a MasterCard can be written as \( A \cap B^c \). We use the formula for the difference of two events: \( P(A \cap B^c) = P(A) - P(A \cap B) \). Substitute the given probabilities: \( P(A \cap B^c) = 0.5 - 0.25 = 0.25 \). Therefore, the probability that the student has a Visa card but not a MasterCard is 0.25.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Union of Events
When we talk about the union of events in probability, we refer to the occurrence of at least one of several events. This concept is essential when we want to determine the likelihood that one event or another happens. In mathematical terms, for two events \( A \) and \( B \), the union is denoted as \( A \cup B \). It gives the probability that either one or both of the events occur.
To calculate the probability of the union of two events \( A \) and \( B \), we use the formula:
  • \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
This formula ensures we do not double-count the intersection鈥攚here both events occur simultaneously鈥攕ince it is included in the probability of \( A \) and \( B \) separately.
In our exercise, for instance, we found that the probability of a student having either a Visa or a MasterCard is 0.65. This result came from adding the probability of having a Visa (0.5) to the probability of having a MasterCard (0.4) and then subtracting the probability of having both (0.25).
Complement Rule
The Complement Rule in probability is a handy technique enabling us to find the probability of an event not occurring. It uses the simple fact that the sum of all probabilities within a certain sample space equals 1.
The complement of an event \( A \), denoted as \( A^c \), represents all outcomes that are not in \( A \). The probability of the complement is calculated as:
  • \( P(A^c) = 1 - P(A) \)
This rule is particularly useful when it is easier to calculate the probability of the complement rather than the event itself.
Referring to our exercise, to find the probability that a student has neither a Visa nor a MasterCard, we calculated the probability of the union of events using \( A \cup B \) first (0.65). The probability of neither event happening is simply the complement, which is 1 minus the probability of \( A \cup B \), resulting in 0.35.
Intersection of Events
The intersection of events in probability theory refers to the scenario where multiple events occur concurrently. For two events \( A \) and \( B \), the intersection is denoted as \( A \cap B \). This symbol represents the event that both \( A \) and \( B \) occur at the same time.
Understanding intersections is crucial when calculating probabilities involving multiple related events. The probability of the intersection of two independent events equals the product of their individual probabilities. However, when events are not independent, information about their joint occurrence鈥擻( P(A \cap B) \)鈥攊s necessary.
In the exercise, the probability of a student having both a Visa card and a MasterCard is given directly as 0.25. Using this, we can derive the probability of only having a Visa by subtracting the intersection probability from the probability of having a Visa alone (0.5), resulting in 0.25 for the probability of having just a Visa and not a MasterCard (\( A \cap B^c \)). This showcases how the intersection helps clarify more complex event scenarios.

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Most popular questions from this chapter

Consider four independent events \(A_{1}, A_{2}, A_{3}\), and \(A_{4}\) and let \(p_{i}=P\left(A_{i}\right)\) for \(i=1,2,3,4\). Express the probability that at least one of these four events occurs in terms of the \(p_{i}\) 's, and do the same for the probability that at least two of the events occur.

A particular airline has 10 a.m. flights from Chicago to New York, Atlanta, and Los Angeles. Let A denote the event that the New York flight is full and define events \(B\) and \(C\) analogously for the other two flights. Suppose \(P(A)=.6, P(B)=.5\), \(P(C)=.4\) and the three events are independent. What is the probability that a. All three flights are full? That at least one flight is not full? b. Only the New York flight is full? That exactly one of the three flights is full?

The College of Science Council has one student representative from each of the five science departments (biology, chemistry, statistics, mathematics, physics). In how many ways can a. Both a council president and a vice president be selected? b. A president, a vice president, and a secretary be selected? c. Two members be selected for the Dean's Council?

Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of time. Consider the two events \(C_{1}=\\{\) left ear tag is lost \(\\}\) and \(C_{2}=\\{\) right ear tag is lost . Let \(\pi=P\left(C_{1}\right)=P\left(C_{2}\right)\), and assume \(C_{1}\) and \(C_{2}\) are independent events. Derive an expression (involving \(\pi\) ) for the probability that exactly one tag is lost given that at most one is lost ("Ear Tag Loss in Red Foxes," J. Wildlife Manag., 1976: 164-167). [Hint: Draw a tree diagram in which the two initial branches refer to whether the left ear tag was lost.]

A chemist is interested in determining whether a certain trace impurity is present in a product. An experiment has a probability of \(.80\) of detecting the impurity if it is present. The probability of not detecting the impurity if it is absent is \(.90\). The prior probabilities of the impurity being present and being absent are \(.40\) and \(.60\), respectively. Three separate experiments result in only two detections. What is the posterior probability that the impurity is present?
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