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Chapter 2: Problem 12

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 Probability of A or B (Union of A and B)

To find the probability of the event \( A \cup B \), use the formula for the probability of the union of two events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]Substitute the given probabilities: \( P(A) = 0.5 \), \( P(B) = 0.4 \), and \( P(A \cap B) = 0.25 \).\[ P(A \cup B) = 0.5 + 0.4 - 0.25 = 0.65 \]
02

Calculate Probability of Neither A nor B

To determine the probability that the selected individual has neither type of card, use the complement rule: \( P((A \cup B)') = 1 - P(A \cup B) \).From Step 1, we have \( P(A \cup B) = 0.65 \).\[ P((A \cup B)') = 1 - 0.65 = 0.35 \]
03

Define and Calculate Probability of a Visa Card Only

The event that the selected student has a Visa card but not a MasterCard is represented by \( A \cap B' \). This is the event that the student has a Visa card but not a MasterCard.To find \( P(A \cap B') \), use the formula: \[ P(A \cap B') = P(A) - P(A \cap B) \]Substitute the given values: \( P(A) = 0.5 \) and \( P(A \cap B) = 0.25 \).\[ P(A \cap B') = 0.5 - 0.25 = 0.25 \]

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Key Concepts

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

Probability Theory
Probability theory is a fundamental branch of mathematics that helps us understand the likelihood of different outcomes. It's all about predicting the future with numbers. By working with probabilities, we can make informed decisions based on expected outcomes rather than mere guesses.
For any event, the probability can range from 0 to 1, where 0 means the event cannot happen, and 1 means it certainly will. The probabilities of all possible outcomes of a random experiment always add up to 1.
The essence of probability theory is to deal with uncertainty in the world. For example, what are the chances it will rain tomorrow? That's where probability steps in to help us calculate and understand these chances better.
Union and Intersection of Events
When dealing with events in probability, it's crucial to understand unions and intersections. - **Union (\(A \cup B\))**: This represents "either A or B or both happening". To calculate the probability of a union of two events, you add their individual probabilities and subtract the probability of both events happening together. The mathematical formula is \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]. Using this formula, we can determine that the likelihood of having at least one type of credit card in the example is 0.65.
- **Intersection (\(A \cap B\))** refers to "both events happening together". In our case, it represents a student having both Visa and MasterCard, with a probability of 0.25.
These operations help us understand more complex scenarios where multiple conditions are involved.
Complement Rule
The complement rule is a handy tool in probability that helps figure out the likelihood of an event 'not' happening. This rule states that the probability of the complement of an event is equal to 1 minus the probability of the event itself. In formula terms, it's expressed as \( P(A') = 1 - P(A) \).
In our example, after determining \( P(A \cup B) = 0.65 \), we use the complement rule to find the probability of neither event happening, represented as \((A \cup B)'\). The calculation goes \[P((A \cup B)') = 1 - 0.65 = 0.35\].
This means there's a 35% chance that a selected student has neither type of credit card. The complement rule is great because it simplifies problems where direct calculation might be more challenging.
Mutually Exclusive Events
Events are called mutually exclusive if they cannot happen at the same time. When two events are mutually exclusive, the probability of both occurring simultaneously is zero, i.e., \( P(A \cap B) = 0 \).
However, in the provided example, we see that \( P(A \cap B) = 0.25 \), which means that A and B are not mutually exclusive. Both can happen simultaneously, as they aren't independent of each other.
Understanding whether events are mutually exclusive can simplify the calculation of probabilities since, for mutually exclusive events, \(P(A \cup B) = P(A) + P(B)\). When they aren鈥檛 mutually exclusive, the intersection probability must be subtracted, as shown in the example.

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Most popular questions from this chapter

Show that if one event \(A\) is contained in another event \(B\) (i.e., \(A\) is a subset of \(B\) ), then \(P(A) \leq P(B)\). [Hint: For such \(A\) and \(B, A\) and \(B \cap A^{\prime}\) are disjoint and \(B=A \cup\left(B \cap A^{\prime}\right)\), as can be seen from a Venn diagram.] For general \(A\) and \(B\), what does this imply about the relationship among \(P(A \cap B), P(A)\) and \(P(A \cup B)\) ?

Each contestant on a quiz show is asked to specify one of six possible categories from which questions will be asked. Suppose \(P(\) contestant requests category \(i)=\frac{1}{6}\) and successive contestants choose their categories independently of one another. If there are three contestants on each show and all three contestants on a particular show select different categories, what is the probability that exactly one has selected category 1 ?

According to the article "Optimization of Distribution Parameters for Estimating Probability of Crack Detection" (J. of Aircraft, 2009: 2090-2097), the following "Palmberg" equation is commonly used to determine the probability \(P_{d}(c)\) of detecting a crack of size \(c\) in an aircraft structure: $$ P_{d}(c)=\frac{\left(c / c^{*}\right)^{\beta}}{1+\left(c / c^{*}\right)^{\beta}} $$ where \(c^{*}\) is the crack size that corresponds to a \(.5\) detection probability (and thus is an assessment of the quality of the inspection process). a. Verify that \(P_{d}\left(c^{*}\right)=.5\) b. What is \(P_{d}\left(2 c^{*}\right)\) when \(\beta=4\) ? c. Suppose an inspector inspects two different panels, one with a crack size of \(c^{*}\) and the other with a crack size of \(2 c^{*}\). Again assuming \(\beta=4\) and also that the results of the two inspections are independent of one another, what is the probability that exactly one of the two cracks will be detected? d. What happens to \(P_{d}(c)\) as \(\beta \rightarrow \infty\) ?

Computer keyboard failures can be attributed to electrical defects or mechanical defects. A repair facility currently has 25 failed keyboards, 6 of which have electrical defects and 19 of which have mechanical defects. a. How many ways are there to randomly select 5 of these keyboards for a thorough inspection (without regard to order)? b. In how many ways can a sample of 5 keyboards be selected so that exactly two have an electrical defect? c. If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of these will have a mechanical defect?

In October, 1994, a flaw in a certain Pentium chip installed in computers was discovered that could result in a wrong answer when performing a division. The manufacturer initially claimed that the chance of any particular division being incorrect was only 1 in 9 billion, so that it would take thousands of years before a typical user encountered a mistake. However, statisticians are not typical users; some modern statistical techniques are so computationally intensive that a billion divisions over a short time period is not outside the realm of possibility. Assuming that the 1 in 9 billion figure is correct and that results of different divisions are independent of one another, what is the probability that at least one error occurs in one billion divisions with this chip?
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