/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Consider randomly selecting a st... [FREE SOLUTION] | 91Ó°ÊÓ

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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’t mutually exclusive, the intersection probability must be subtracted, as shown in the example.

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Most popular questions from this chapter

A factory uses three production lines to manufacture cans of a certain type. The accompanying table gives percentages of nonconforming cans, categorized by type of nonconformance, for each of the three lines during a particular time period. $$ \begin{array}{lccc} \hline & \text { Line 1 } & \text { Line 2 } & \text { Line 3 } \\ \hline \text { Blemish } & 15 & 12 & 20 \\ \text { Crack } & 50 & 44 & 40 \\ \text { Pull-Tab Problem } & 21 & 28 & 24 \\ \text { Surface Defect } & 10 & 8 & 15 \\ \text { Other } & 4 & 8 & 2 \\ \hline \end{array} $$ During this period, line 1 produced 500 nonconforming cans, line 2 produced 400 such cans, and line 3 was responsible for 600 nonconforming cans. Suppose that one of these 1500 cans is randomly selected. a. What is the probability that the can was produced by line 1? That the reason for nonconformance is a crack? b. If the selected can came from line 1 , what is the probability that it had a blemish? c. Given that the selected can had a surface defect, what is the probability that it came from line 1?

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 ?

A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift. A quality control consultant is to select 6 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 6 workers has the same chance of being selected as does any other group (drawing 6 slips without replacement from among 45). a. How many selections result in all 6 workers coming from the day shift? What is the probability that all 6 selected workers will be from the day shift? b. What is the probability that all 6 selected workers will be from the same shift? c. What is the probability that at least two different shifts will be represented among the selected workers? d. What is the probability that at least one of the shifts will be unrepresented in the sample of workers?

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehicles pass or fail independently of one another, calculate the following probabilities: a. \(P\) (all of the next three vehicles inspected pass) b. \(P\) (at least one of the next three inspected fails) c. \(P\) (exactly one of the next three inspected passes) d. \(P\) (at most one of the next three vehicles inspected passes) e. Given that at least one of the next three vehicles passes inspection, what is the probability that all three pass (a conditional probability)?

The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop machines and the other four have chosen desktop machines. Suppose that only two of the setups can be done on a particular day, and the two computers to be set up are randomly selected from the six (implying 15 equally likely outcomes; if the computers are numbered \(1,2, \ldots, 6\), then one outcome consists of computers 1 and 2 , another consists of computers 1 and 3 , and so on). a. What is the probability that both selected setups are for laptop computers? b. What is the probability that both selected setups are desktop machines? c. What is the probability that at least one selected setup is for a desktop computer? d. What is the probability that at least one computer of each type is chosen for setup?

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