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Chapter 4: Problem 77

Given that \(P(B)=.65\) and \(P(A\) and \(B)=.45\), find \(P(A \mid B)\).

Short Answer

Expert verified
The conditional probability \(P(A \mid B)\) is approximately 0.692.

Step by step solution

01

Identify the Given Probabilities

From the problem, we know that \(P(B) = .65\) and \(P(A \text{ and } B) = .45\).
02

Apply the Conditional Probability Formula

Substitute the given probabilities into the formula for conditional probability, \(P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)}\). This gives us \(P(A \mid B) = \frac{.45}{.65}\).
03

Evaluate the Expression

Finally, calculate the value of \(P(A \mid B)\). It's approximately 0.692.

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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 branch of mathematics focused on the analysis of random events and outcomes. It provides the tools to quantify uncertainty and reason about the likelihood of different events. Understanding probability theory is crucial because it serves as the foundation for statistics, helping us to make informed decisions based on uncertain situations.
  • Probabilities range from 0 to 1, where 0 indicates impossibility and 1 indicates certainty.
  • An event is a set outcome that may or may not occur, and probability measures the likelihood of this event.
  • In this context, the probability of an event A, denoted as \(P(A)\), is calculated based on the relative frequency of the event occurring out of all possible outcomes.
Grasping these basics allows us to dive into more complex ideas like joint probability and conditional probability. These concepts are often interrelated and constitute the backbone for solving real-world problems involving random variables.
Joint Probability
Joint probability refers to the probability of two events happening at the same time. It's a fundamental concept that allows us to explore the likelihood of two or more events occurring together. Joint probability is denoted as \(P(A \text{ and } B)\) or sometimes \(P(A \cap B)\), which reads as "the probability of A and B."
  • This concept is especially useful when analyzing scenarios where multiple factors are at play.
  • The joint probability is calculated differently depending on whether events are independent or dependent.
  • In the context of dependent events, the joint probability can be thought of as the product of the conditional probability \(P(A \mid B)\) and \(P(B)\).
This foundational understanding of joint probabilities is key to solving more complex probability puzzles and helps in understanding how events influence each other.
Probability Formulas
When solving probability problems, having a strong grasp of probability formulas can be incredibly helpful. These formulas allow us to calculate various types of probabilities, including joint, marginal, and conditional probabilities.
To solve the given exercise, we used the formula for conditional probability:
  • \(P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)}\)
This formula indicates that the conditional probability \(P(A \mid B)\) is computed by dividing the joint probability \(P(A \text{ and } B)\) by the probability of the event B \(P(B)\).
  • Conditional probability is essential when computing the probability of an event occurring, given that another event has already occurred.
  • These formulas are interconnected, and understanding one often leads to a better comprehension of others, enhancing overall problem-solving skills.
The efficient application of these probability formulas is crucial for research, data analysis, and many real-life applications where outcomes are uncertain.

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Most popular questions from this chapter

In a sample survey, 1800 senior citizens were asked whether or not they have ever been victimized by a dishonest telemarketer. The following table gives the responses by age group (in years). $$\begin{array}{l|lccc} & & & \begin{array}{c} \text { Have Been } \\ \text { Victimized } \end{array} & \begin{array}{c} \text { Have Never } \\ \text { Been Victimized } \end{array} \\ \hline & 60-69 & \text { (A) } & 106 & 698 \\ \text { Age } & 70-79 & \text { (B) } & 145 & 447 \\ & 80 \text { or over } &\text { (C) } & 61 & 343 \\ \hline \end{array}$$ a. Suppose one person is randomly selected from these senior citizens. Find the following probabilities. i. \(P(\) have been victimized and \(\mathrm{C}\) ) ii. \(P(\) have never been victimized and \(\mathrm{A}\) ) b. Find \(P(B\) and \(C\) ). Is this probability zero? Explain why or why not.

A car rental agency currently has 44 cars available, 28 of which have a GPS navigation system. One of the 44 cars is selected at random. Find the probability that this car a. has a GPS navigation system b. does not have a GPS navigation system

A gambler has given you two jars and 20 marbles. Of these 20 marbles, 10 are red and 10 are green You must put all 20 marbles in these two jars in such a way that each jar must have at least one marble in it. Then a friend of yours, who is blindfolded, will select one of the two jars at random and then will randomly select a marble from this jar. If the selected marble is red, you and your friend win \(\$ 100\) a. If you put 5 red marbles and 5 green marbles in each jar, what is the probability that your friend selects a red marble? b. If you put 2 red marbles and 2 green marbles in one jar and the remaining marbles in the other jar, what is the probability that your friend selects a red marble? c. How should these 20 marbles be distributed among the two jars in order to give your friend the highest possible probability of selecting a red marble?

Many states have a lottery game, usually called a Pick-4, in which you pick a four-digit number such as 7359 . During the lottery drawing, there are four bins, each containing balls numbered 0 through 9\. One ball is drawn from each bin to form the four-digit winning number. a. You purchase one ticket with one four-digit number. What is the probability that you will win this lottery game? b. There are many variations of this game. The primary variation allows you to win if the four digits in your number are selected in any order as long as they are the same four digits as obtained by the lottery agency. For example, if you pick four digits making the number 1265, then you will win if \(1265,2615,5216,6521\), and so forth, are drawn. The variations of the lottery game depend on how many unique digits are in your number. Consider the following four different versions of this game. i. All four digits are unique (e.g., 1234 ) ii. Exactly one of the digits appears twice (e.g., 1223 or 9095 ) iii. Two digits each appear twice (e.g., 2121 or 5588 ) iv. One digit appears three times (e.g., 3335 or 2722 ) Find the probability that you will win this lottery in each of these four situations.

A random sample of 400 college students was asked if college athletes should be paid. The following table gives a two-way classification of the responses. $$\begin{array}{lcc} \hline & \text { Should Be Paid } & \text { Should Not Be Paid } \\ \hline \text { Student athlete } & 90 & 10 \\ \text { Student nonathlete } & 210 & 90 \end{array}$$ a. If one student is randomly selected from these 400 students, find the probability that this student i. is in favor of paying college athletes ii. favors paying college athletes given that the student selected is a nonathlete iii. is an athlete and favors paying student athletes iv. is a nonathlete or is against paying student athletes b. Are the events "student athlete" and "should be paid" independent? Are they mutually exclusive? Explain why or why not.
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