/*! 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 70 Given that \(P(A)=.72\) and \(P(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Given that \(P(A)=.72\) and \(P(A\) and \(B)=.38\), find \(P(B \mid A)\).

Short Answer

Expert verified
The conditional probability \(P(B|A)\) is approximately \(0.527778\). Please, note that this value may vary slightly depending on the rounding-off technique used.

Step by step solution

01

Identify Known And Required Variables

From the problem, the known variables are: \(P(A) = 0.72\) and \(P(A \text{ and } B) = 0.38\). The required unknown to find is \(P(B|A)\), the probability of event B occurring given event A has already occurred.
02

Apply Conditional Probability Formula

The formula for calculating conditional probability is given as \(P(B|A) = \frac{{P(A \text{ and } B)}}{{P(A)}\). Applying this formula with the known variables is the next step.
03

Substitute Known Variables Into the Formula

Substitute \(P(A) = 0.72\) and \(P(A \text{ and } B) = 0.38\) into the formula to derive \(P(B|A) = \frac{0.38}{0.72}\).
04

Compute the Value of \(P(B|A)\)

Divide 0.38 by 0.72 to compute the value of \(P(B|A)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 that deals with the likelihood of events occurring. It allows us to assign values to uncertain outcomes. These outcomes can be as simple as flipping a coin or as complex as predicting weather patterns.

In probability theory, we work with events, which are outcomes or combinations of outcomes. An event's probability is a value between 0 and 1. A probability of 0 means the event will not occur, while a probability of 1 indicates certainty.
  • For example, the probability of rolling a 4 on a standard six-sided die is \( \frac{1}{6} \).
  • Probabilities can be subjective or objective based on frequency or logical analysis.
Understanding probability theory is crucial for analyzing different scenarios where uncertainty is present.
The analysis and predictions derived from probability theory form the backbone of statistical inference.
Statistics
Statistics is the science of collecting, analyzing, and interpreting data. It's used in a wide range of fields to make informed decisions based on data.

Statistics involves various types of analysis, including descriptive statistics, which summarize data collected for a study, and inferential statistics, which draw conclusions from data. Here are key elements:
  • Descriptive Statistics: Include mean, median, mode, and standard deviation. These help summarize the data.
  • Inferential Statistics: Used to infer trends about a larger population from samples. This involves probability theory heavily, using concepts like confidence intervals and hypothesis testing.
Statistics is not only about calculations but also about understanding the data and extracting meaningful insights from it. Overall, it's a critical tool for decision making in uncertain environments.
Event Intersection
Event Intersection involves looking at the probability of two or more events happening simultaneously. This is often represented by the intersection symbol, typically noted as \(P(A \text{ and } B)\).

The intersection of events is pivotal in computing probabilities like those seen in conditional probability. Let's consider:
  • When calculating \(P(A \text{ and } B)\), you are determining the likelihood for both event A and event B to happen at the same time.
  • The relationship between event intersection and conditional probability is essential for problems where one event depends on another.
For instance, if you know \(P(A \text{ and } B)\) and \(P(A)\), you can use the formula \(P(B|A) = \frac{P(A \text{ and } B)}{P(A)}\) to find the probability of B given A.
Event intersection provides the critical underpinning for understanding how events relate to each other within the realm of probability.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

How is the multiplication rule of probability for two dependent events different from the rule for two independent events?

A production system has two production lines; each production line performs a two-part process, and each process is completed by a different machine. Thus, there are four machines, which we can identify as two first-level machines and two second-level machines. Each of the first-level machines works properly \(98 \%\) of the time, and each of the second-level machines works properly \(96 \%\) of the time. All four machines are independent in regard to working properly or breaking down. Two products enter this production system, one in each production line. a. Find the probability that both products successfully complete the two-part process (i.e., all four machines are working properly). b. Find the probability that neither product successfully completes the two- part process (i.e., at least one of the machines in each production line is not working properly).

Two thousand randomly selected adults were asked if they think they are financially better off than their parents. The following table gives the two- way classification of the responses based on the education levels of the persons included in the survey and whether they are financially better off, the same as. or worse off than their parents. $$ \begin{array}{lccc} \hline & \begin{array}{c} \text { Less Than } \\ \text { High School } \end{array} & \begin{array}{c} \text { High } \\ \text { School } \end{array} & \begin{array}{c} \text { More Than } \\ \text { High School } \end{array} \\ \hline \text { Better off } & 140 & 450 & 420 \\ \text { Same as } & 60 & 250 & 110 \\ \text { Worse off } & 200 & 300 & 70 \\ \hline \end{array} $$ Suppose one adult is selected at random from these 2000 adults. Find the following probabilities. a. \(P\) (better off or high school) b. \(P\) (more than high school or worse off) c. \(P(\) better off or worse off \()\)

Given that \(P(B)=.29\) and \(P(A\) and \(B)=.24\), find \(P(A \mid B)\).

Given that \(A\) and \(B\) are two mutually exclusive events, find \(P(A\) or \(B\) ) for the following. a. \(P(A)=.71\) and \(P(B)=.03\) b. \(P(A)=.44\) and \(P(B)=.38\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.