# Compute $P\left( A|B \right)$, if $P\left( B \right)=0.25$ and $P\left( A\cap B \right)=0.18$.

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Hint: In probability, we have a formula $P\left( A|B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$. This formula can be used to solve this question.

Before proceeding with the question, we must know all those formulas of probability that can be used to solve this question. There is only one formula which can be used to solve this question. That formula is,

$P\left( A|B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}..............\left( 1 \right)$

Here, $A$ and $B$ are the two mutually exclusive events and $P\left( A|B \right)$ represents the probability of event $A$ given that the event $B$ will also take place. $P\left( A\cap B \right)$ represents the probability that both the events will occur simultaneously.

In this question, we are given some information on the probability related to event $A$ and event $B$. We are given,

$P\left( B \right)=0.25$

$P\left( A\cap B \right)=0.18$

In the question, we are required to find the probability $P\left( A|B \right)$. It can be seen from equation $\left( 1 \right)$ that we require $P\left( B \right)$ and $P\left( A\cap B \right)$ to find the value of $P\left( A|B \right)$. In this question, we are given both $P\left( B \right)$ and $P\left( A\cap B \right)$. So, substituting $P\left( B \right)=0.25$ and $P\left( A\cap B \right)=0.18$ in equation $\left( 1 \right)$, we get,

$\begin{align}

& P\left( A|B \right)=\dfrac{0.18}{0.25} \\

& \Rightarrow P\left( A|B \right)=0.72 \\

\end{align}$

So, the value of $P\left( A|B \right)$ is equal to $0.72$.

Note: There is a possibility that one make commit a mistake while using the formula to find $P\left( A|B \right)$ i.e. $P\left( A|B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$. Sometimes instead of using the formula $P\left( A|B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$, we use the formula $P\left( A|B \right)=\dfrac{P\left( A\cap B \right)}{P\left( A \right)}$. So, instead of substituting the value of $P\left( B \right)$, we find the value of $P\left( A \right)$ by using the formula $P\left( A \right)=1-P\left( B \right)$ and substitute this value of $P\left( A \right)$ in the formula $P\left( A|B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$. This mistake may lead us to the incorrect value of $P\left( A|B \right)$ and eventually, we end up marking the incorrect option if the question is a multiple choice question or give incorrect answers if the question is a subjective one.

Before proceeding with the question, we must know all those formulas of probability that can be used to solve this question. There is only one formula which can be used to solve this question. That formula is,

$P\left( A|B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}..............\left( 1 \right)$

Here, $A$ and $B$ are the two mutually exclusive events and $P\left( A|B \right)$ represents the probability of event $A$ given that the event $B$ will also take place. $P\left( A\cap B \right)$ represents the probability that both the events will occur simultaneously.

In this question, we are given some information on the probability related to event $A$ and event $B$. We are given,

$P\left( B \right)=0.25$

$P\left( A\cap B \right)=0.18$

In the question, we are required to find the probability $P\left( A|B \right)$. It can be seen from equation $\left( 1 \right)$ that we require $P\left( B \right)$ and $P\left( A\cap B \right)$ to find the value of $P\left( A|B \right)$. In this question, we are given both $P\left( B \right)$ and $P\left( A\cap B \right)$. So, substituting $P\left( B \right)=0.25$ and $P\left( A\cap B \right)=0.18$ in equation $\left( 1 \right)$, we get,

$\begin{align}

& P\left( A|B \right)=\dfrac{0.18}{0.25} \\

& \Rightarrow P\left( A|B \right)=0.72 \\

\end{align}$

So, the value of $P\left( A|B \right)$ is equal to $0.72$.

Note: There is a possibility that one make commit a mistake while using the formula to find $P\left( A|B \right)$ i.e. $P\left( A|B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$. Sometimes instead of using the formula $P\left( A|B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$, we use the formula $P\left( A|B \right)=\dfrac{P\left( A\cap B \right)}{P\left( A \right)}$. So, instead of substituting the value of $P\left( B \right)$, we find the value of $P\left( A \right)$ by using the formula $P\left( A \right)=1-P\left( B \right)$ and substitute this value of $P\left( A \right)$ in the formula $P\left( A|B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$. This mistake may lead us to the incorrect value of $P\left( A|B \right)$ and eventually, we end up marking the incorrect option if the question is a multiple choice question or give incorrect answers if the question is a subjective one.

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