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It is given that the events \(A\) and \(B\) are such that \(P(A)=\frac{1}{4}, P(A \mid B)=\frac{1}{2}\) and \(P(B \mid A)=\frac{2}{3} .\) Then \(P(B)\) is [2008] (a) \(\frac{1}{6}\) (b) \(\frac{1}{3}\) (c) \(\frac{2}{3}\) (d) \(\frac{1}{2}\)

Short Answer

Expert verified
\(P(B) = \frac{1}{3}\), option (b).

Step by step solution

01

Understanding Given Probabilities

We are provided with three probabilities: \(P(A) = \frac{1}{4}\), \(P(A \mid B) = \frac{1}{2}\), and \(P(B \mid A) = \frac{2}{3}\). These are known as conditional probabilities and direct probabilities.
02

Use Bayes' Theorem

To find \(P(B)\), start with Bayes' Theorem: \(P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}\). We will rearrange this formula later to solve for \(P(B)\).
03

Rearrange Bayes' Theorem

Rearrange the formula from Step 2 to express \(P(B)\) in terms of the known probabilities.\[ P(B) = \frac{P(B \mid A) \cdot P(A)}{P(A \mid B)} \]
04

Substitute Given Values

Substitute the known values into the rearranged formula:\[ P(B) = \frac{\frac{2}{3} \times \frac{1}{4}}{\frac{1}{2}} \]
05

Perform the Calculations

First, calculate the numerator: \(\frac{2}{3} \times \frac{1}{4} = \frac{2}{12} = \frac{1}{6}\).Next, divide by the denominator \(\frac{1}{2}\):\[ P(B) = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{6} \times 2 = \frac{1}{3} \]
06

Consolidate the Solution

Thus, the probability \(P(B)\) is \(\frac{1}{3}\).

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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 that describes the likelihood of an event occurring given that another event has already occurred. It's all about using the additional information to update the probability of an event.
For example, in the exercise, we have the conditional probabilities \(P(A \mid B) = \frac{1}{2}\) and \(P(B \mid A) = \frac{2}{3}\). Here, \(P(A \mid B)\) means "the probability of event \(A\) occurring given that \(B\) has already occurred."
To calculate any conditional probability, use the formula:
\[P(A \mid B) = \frac{P(A \cap B)}{P(B)}\]where \(P(A \cap B)\) is the probability that both \(A\) and \(B\) happen simultaneously. Understanding this concept helps in making more informed probability calculations when dealing with dependent events.
Probability Calculation
Probability calculation is at the heart of solving probability problems. It involves using known values and mathematical formulas to find unknown probabilities. In the given exercise, we applied probability calculations in several steps.
First, we identified that we could use Bayes' Theorem to find the missing probability \(P(B)\). This is a common approach when dealing with problems involving conditional probabilities. The relevant formula from Bayes' Theorem is:
\[ P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)} \] By rearranging, it becomes:
\[ P(B) = \frac{P(B \mid A) \cdot P(A)}{P(A \mid B)} \]
Substituting the known values into this equation allows us to compute \(P(B)\) using straightforward arithmetic operations. This step-by-step calculation illustrates how systematic probability calculations help us derive unknown probabilities from known pieces of information.
Probability Theory
Probability theory provides the framework and rules for dealing with random events and their likelihoods. It allows us to quantify how likely events are to happen and make predictions based on mathematical principles. This theory encompasses various concepts such as events, outcomes, and probability measures.
In the exercise, probability theory helps us understand how probabilities are interrelated, such as how the probability of \(B\) depends on the given probabilities of \(A\) and the conditional probabilities \(P(A \mid B)\) and \(P(B \mid A)\).
  • **Events**: Basic outcomes like \(A\) and \(B\), with each having specified probabilities.
  • **Probability Measures**: Quantifying the chance of an event's occurrence, like \(P(A) = \frac{1}{4}\).
  • **Bayes' Theorem**: A pivotal rule in probability theory that relates conditional probabilities and offers a method to compute unknown probabilities.
Understanding probability theory is crucial for interpreting and solving complex probability problems, ensuring predictions are accurate and formulas are correctly applied.

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Most popular questions from this chapter

The probability that a randomly chosen 5 -digit number is made from exactly two digits is: \(\quad\) [Sep.03, 2020 (II)] (a) \(\frac{135}{10^{4}}\) (b) \(\frac{121}{10^{4}}\) (c) \(\frac{150}{10^{4}}\) (d) \(\frac{134}{10^{4}}\)

A random variable \(X\) has Poisson distribution with mean \(2 .\) Then \(P(X>1.5)\) equals (a) \(\frac{2}{e^{2}}\) (b) 0 (c) \(1-\frac{3}{e^{2}}\) (d) \(\frac{3}{e^{2}}\)

Let \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) be three events, which are pair- wise independence and \(\bar{E}\) denotes the complement of an event E. If \(\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0\) and \(\mathrm{P}(\mathrm{C})>0\), then \(\mathrm{P}[(\bar{A} \cap \bar{B}) \mid \mathrm{C}]\) is equal to. \(\quad\) Online April 16, 2018] (a) \(\mathrm{P}(\mathrm{A})+\mathrm{P}(\bar{B})\) (b) \(\mathrm{P}(\bar{A})-\mathrm{P}(\bar{B})\) (c) \(\mathrm{P}(\bar{A})-\mathrm{P}(\mathrm{B})\) (d) \(\mathrm{P}(\bar{A})+\mathrm{P}(\bar{B})\)

A player \(X\) has a biased coin whose probability of showing heads is \(p\) and a player \(Y\) has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If \(X\) starts the game, and the probability of winning the game by both the players is equal, then the value of ' \(p\) ' is[Online April 15, 2018] (a) \(\frac{1}{3}\) (b) \(\frac{1}{5}\) (c) \(\frac{1}{4}\) (d) \(\frac{2}{5}\)

Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than \(99 \%\) is : \(\quad\) [April 10, 2019 (II)] (a) 5 (b) 6 (c) 8 (d) 7

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