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Suppose that \(P(A \mid B)=0.7, P(A)=0.5\), and \(P(B)=\) 0.2. Determine \(P(B \mid A)\)

Short Answer

Expert verified
The probability \( P(B \mid A) \) is 0.28.

Step by step solution

01

Understanding the Problem

We are given the conditional probability \( P(A \mid B) = 0.7 \), the probability \( P(A) = 0.5 \), and \( P(B) = 0.2 \). We need to find the probability \( P(B \mid A) \). To do this, we'll use Bayes' theorem, which connects these probabilities.
02

Applying Bayes' Theorem

Bayes' theorem states: \[ P(B \mid A) = \frac{P(A \mid B) \cdot P(B)}{P(A)} \]. We will substitute the known values: \( P(A \mid B) = 0.7 \), \( P(B) = 0.2 \), and \( P(A) = 0.5 \) into this formula.
03

Substituting the Values

Plug in the values into Bayes' theorem formula: \[ P(B \mid A) = \frac{0.7 \times 0.2}{0.5} \]. This calculation will give us the probability of \( B \) given \( A \).
04

Calculating the Numerator

Calculate the numerator of the fraction: \( 0.7 \times 0.2 = 0.14 \).
05

Calculating the Final Result

Now calculate \( \frac{0.14}{0.5} = 0.28 \). This is the probability \( P(B \mid A) \).

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Key Concepts

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

Conditional Probability
Conditional probability is a fundamental concept in probability theory. It refers to the probability of an event occurring given that another event has already occurred. This helps us understand the likelihood of one event in relation to a known condition.

Let's consider the example: suppose we know that event B has occurred, and we want to find the probability of event A happening given this information. This is denoted as \( P(A \mid B) \). The formula for conditional probability is:
  • \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \)
Here, \( P(A \cap B) \) is the probability of both events A and B happening, while \( P(B) \) is the probability of event B.

Understanding conditional probability is crucial because it allows us to update our predictions based on new, relevant information. This is used in various fields, like risk assessment and decision making, where conditions constantly change.
Probability Theory
Probability theory is the branch of mathematics concerned with the analysis of random phenomena. It provides a framework for understanding events governed by uncertainty. The foundation of probability theory includes basic concepts like events, sample spaces, and probability measures.

Each outcome in a sample space has a likelihood, which is expressed as a probability value ranging from 0 to 1.
  • 0 indicates an impossible event.
  • 1 indicates a certain event.
In probability theory, we use rules and formulas to compute complex probabilities, like in Bayes' theorem. Bayes' theorem allows you to find a probability when you know certain other probabilities related to an event.

These principles apply in everyday situations such as predicting weather changes, stock market behaviors, or simply rolling a dice. Probability theory helps us manage and interpret every day's inherent randomness.
Applied Statistics
Applied statistics involves the use of statistical techniques to solve real-world problems by collecting, analyzing, and interpreting data. It enables us to draw informed conclusions based on data evaluation.

Statistics use probability as a key tool to understand patterns and relationships. In the context of statistics:
  • Probabilities help identify trends and anomalies.
  • They support the validation of models used to represent data.
Through applied statistics, we're empowered to make predictions and decisions based on historical data. This is particularly valuable in fields such as marketing, finance, healthcare, and engineering where data-driven decisions can lead to significant benefits.

In practice, applied statistics turn probability theory and statistical formulas, like Bayes' theorem, into actionable insights, shaping strategies and informing policies.

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Most popular questions from this chapter

The sample space of a random experiment is \(\\{a, b, c\) \(d, e\\}\) with probabilities \(0.1,0.1,0.2,0.4,\) and \(0.2,\) respectively. Let \(A\) denote the event \(\\{a, b, c\\},\) and let \(B\) denote the event \(\\{c, d, e\\} .\) Determine the following: a. \(P(A)\) b. \(P(B)\) c. \(P\left(A^{\prime}\right)\) d. \(P(A \cup B)\) e. \(P(A \cap B)\)

An article in The Canadian Entomologist (Harcourt et al., 1977 , Vol. 109 , pp. \(1521-1534\) ) reported on the life of the alfalfa weevil from eggs to adulthood. The following table shows the number of larvae that survived at each stage of development from eggs to adults. $$ \begin{array}{cccccc} & \text { Early } & \text { Late } & \text { Pre- } & \text { Late } & \\ \text { Eggs } & \text { Larvae } & \text { Larvae } & \text { pupae } & \text { Pupae } & \text { Adults } \\ 421 & 412 & 306 & 45 & 35 & 31 \end{array} $$ a. What is the probability an egg survives to adulthood? b. What is the probability of survival to adulthood given survival to the late larvae stage? c. What stage has the lowest probability of survival to the next stage?

2.4.11 The article "Clinical and Radiographic Outcomes of Four Different Treatment Strategies in Patients with Early Rheumatoid Arthritis" [Arthritis \& Rheumatism (2005, Vol. 52, pp. \(3381-3390\) ) ] considered four treatment groups. The groups consisted of patients with different drug therapies (such as prednisone and infliximab): sequential monotherapy (group 1), step-up combination therapy (group 2), initial combination therapy (group 3), or initial combination therapy with infliximab (group 4). Radiographs of hands and feet were used to evaluate disease progression. The number of patients without progression of joint damage was 76 of 114 patients \((67 \%), 82\) of 112 patients \((73 \%), 104\) of 120 patients \((87 \%),\) and 113 of 121 patients \((93 \%)\) in groups \(1-4\) respectively. Suppose that a patient is selected randomly. Let \(A\) denote the event that the patient is in group \(1,\) and let \(B\) denote the event that there is no progression. Determine the following probabilities: a. \(P(A \cup B)\) b. \(P\left(A^{\prime} \cup B^{\prime}\right)\) c. \(P\left(A \cup B^{\prime}\right)\)

A credit card contains 16 digits. It also contains the month and year of expiration. Suppose there are 1 million credit card holders with unique card numbers. A hacker randomly selects a 16-digit credit card number. a. What is the probability that it belongs to a user? b. Suppose a hacker has a \(25 \%\) chance of correctly guessing the year your card expires and randomly selects 1 of the 12 months. What is the probability that the hacker correctly selects the month and vear of expiration?

A computer system uses passwords that contain exactly eight characters, and each character is one of the 26 lowercase letters \((a-z)\) or 26 uppercase letters \((A-Z)\) or 10 integers \((0-9)\) Let \(\Omega\) denote the set of all possible passwords. Suppose that all passwords in \(\Omega\) are equally likely. Determine the probability for each of the following: a. Password contains all lowercase letters given that it contains only letters b. Password contains at least 1 uppercase letter given that it contains only letters c. Password contains only even numbers given that it contains all numbers

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