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Chapter 3: Problem 17

In a certain community, 36 percent of the families own a dog and 22 percent of the families that own a dog also own a cat. In addition, 30 percent of the families own a cat. What is (a) the probability that a randomly selected family owns both a dog and a cat? (b) the conditional probability that a randomly selected family owns a dog given that it owns a cat?

Short Answer

Expert verified
(a) The probability that a randomly selected family owns both a dog and a cat is 0.0792 or 7.92%. (b) The conditional probability that a randomly selected family owns a dog given that it owns a cat is 0.264 or 26.4%.

Step by step solution

01

(Step 1: Finding the joint probability of owning a dog and a cat)

The problem states that 22 percent of the families that own a dog also own a cat. Since 36 percent of the families own a dog, we can find the joint probability of owning a dog and a cat by multiplying the percentage of families that own a dog by the percentage of dog owners who also own a cat: Joint probability = \(0.36 * 0.22 = 0.0792\) So, 7.92 percent of the families own both a dog and a cat.
02

(Step 1a: Answer for part (a))

The probability that a randomly selected family owns both a dog and a cat is equal to the joint probability we just found, which is: P(A and B) = 0.0792 (a) Answer: The probability that a randomly selected family owns both a dog and a cat is 0.0792 or 7.92%.
03

(Step 2: Finding the conditional probability of owning a dog given the family owns a cat)

To find the conditional probability of owning a dog given the family owns a cat, we will use the formula for conditional probability: P(Dog | Cat) = \(\frac{P(Dog \cap Cat)}{P(Cat)}\) We already found the joint probability P(Dog and Cat) and we know that 30 percent of the families own a cat. So, we can compute the conditional probability as: \(P(Dog | Cat) = \frac{0.0792}{0.30} = 0.264\)
04

(Step 2a: Answer for part (b))

The conditional probability that a randomly selected family owns a dog given that it owns a cat is: P(Dog | Cat) = 0.264 (b) Answer: The conditional probability that a randomly selected family owns a dog given that it owns a cat is 0.264 or 26.4%.

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Key Concepts

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

Joint Probability
Understanding joint probability is a foundational aspect of probability theory. It refers to a measure of two events occurring at the same time. In our exercise, we calculate the joint probability of a family owning both a dog and a cat. This is done by multiplying the individual probabilities of the two independent events鈥攐wning a dog and owning a cat given that the family owns a dog. It's essential to realize that calculating joint probability is only this straightforward when the events are independent, meaning the occurrence of one does not affect the occurrence of the other. In context, the joint probability helped us answer part (a) of the exercise by revealing that 7.92% of all families own both a dog and a cat.
Probability Theory
Probability theory is the branch of mathematics concerned with the analysis of random phenomena. The central objects of probability theory are random variables, stochastic processes, and events: mathematical abstractions of non-deterministic events or measured quantities that may either occur single or evolve over time in an apparently random fashion. In the exercise, we delve into this fascinating area by looking at how likely certain events are鈥攕uch as a family owning a pet. Probability theory employs various formulas, including those used for joint and conditional probabilities, to quantify the likelihood of events and to make informed predictions based on statistical data. Grasping the basic principles of probability theory enables students to solve complex real-world problems that involve uncertainty and variability.
Conditional Probability Formula
Conditional probability is the likelihood of an event occurring given that another event has already occurred. The conditional probability formula is elegantly simple but immensely powerful: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\), where \(P(A|B)\) represents the probability of event A occurring given that B is true. In the exercise, we used this formula to determine the probability of a family owning a dog given they own a cat鈥攊llustrating it with real numbers to answer part (b) of the problem. One crucial point to remember is that conditional probability can profoundly change our expectations regarding the occurrence of an event, as it incorporates additional information that might not be considered in a simple probability calculation. It's a vital concept for students to understand, as it's widely used in various fields, from risk assessment to machine learning algorithms.

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Most popular questions from this chapter

If you had to construct a mathematical model for events \(E\) and \(F,\) as described in parts (a) through (e), would you assume that they were independent events? Explain your reasoning. (a) \(E\) is the event that a businesswoman has blue eyes, and \(F\) is the event that her secretary has blue eyes. (b) \(E\) is the event that a professor owns a car, and \(F\) is the event that he is listed in the telephone book. (c) \(E\) is the event that a man is under 6 feet tall, and \(F\) is the event that he weighs more than 200 pounds. (d) \(E\) is the event that a woman lives in the United States, and \(F\) is the event that she lives in the Western Hemisphere. (e) \(E\) is the event that it will rain tomorrow, and \(F\) is the event that it will rain the day after tomorrow.

Independent flips of a coin that lands on heads with probability \(p\) are made. What is the probability that the first four outcomes are (a) \(H, H, H, H ?\) (b) \(T, H, H, H ?\) (c) What is the probability that the pattern \(T, H, H, H\) occurs before the pattern \(H, H, H, H ?\) Hint for part \((c):\) How can the pattern \(H, H, H, H\) occur first?

Suppose that an insurance company classifies people into one of three classes: good risks, average risks, and bad risks. The company's records indicate that the probabilities that good-, average-, and bad-risk persons will be involved in an accident over a 1 -year span are, respectively, .05, .15 and \(30 .\) If 20 percent of the population is a good risk, 50 percent an average risk, and 30 percent a bad risk, what proportion of people have accidents in a fixed year? If policyholder \(A\) had no accidents in \(2012,\) what is the probability that he or she is a good risk? is an average risk?

Barbara and Dianne go target shooting. Suppose that each of Barbara's shots hits a wooden duck target with probability \(p_{1},\) while each shot of Dianne's hits it with probability \(p_{2} .\) Suppose that they shoot simultaneously at the same target. If the wooden duck is knocked over (indicating that it was hit), what is the probability that (a) both shots hit the duck? (b) Barbara's shot hit the duck? What independence assumptions have you made?

Suppose that we want to generate the outcome of the flip of a fair coin, but that all we have at our disposal is a biased coin that lands on heads with some unknown probability \(p\) that need not be equal to \(\frac{1}{2} .\) Consider the following procedure for accomplishing our task: 1\. Flip the coin. 2\. Flip the coin again. 3\. If both flips land on heads or both land on tails, return to step 1. 4\. Let the result of the last flip be the result of the experiment. (a) Show that the result is equally likely to be either heads or tails. (b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different and then lets the result be the outcome of the final flip?
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