# If $P(A)=0.4,P(B)=0.8,P\left( \dfrac{B}{A} \right)=0.6$ . Find $P\left( \dfrac{A}{B} \right)$ .

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Hint: This question is based on conditional probability . We have to find the probability of occurrence of event \[A\] with a condition that event $B$ has already occurred.

Before we proceed with the solution , we must know the concept of conditional probability.

When the probability of happening of one event is affected by the happening of another, then the two events are known as dependent events.

For example : There are $2$ black cards and $5$ red cards in a pile. The probability of picking a red card is given as $\dfrac{5}{7}$ . But , after removing the red card , now the probability of picking a red or a black card changes . The probability of picking a red card becomes $\dfrac{4}{6}=\dfrac{2}{3}$ and the probability of picking a black card becomes $\dfrac{2}{6}=\dfrac{1}{3}$ .

Now , if $A$ and \[B\] are two dependent events , then the probability of happening of $A$ given that \[B\] has already happened is given by $P\left( \dfrac{A}{B} \right)=\dfrac{P(A\cap B)}{P(B)}$ , where $P(B)$ is the probability of happening of event \[B\] and $P(A\cap B)$ is the probability of happening of events $A$ and \[B\] simultaneously.

Now , coming to the question , we are given probability of happening of event $A$ is equal to $0.4$ , the probability of happening of event $B$ is equal to $0.8$ and the probability of happening of the event $B$ given that $A$ has happened is $0.6$ .

Now ,from the information that the probability of happening of the event $B$ given that $A$ has happened is $0.6$, we can write $\dfrac{P(A\cap B)}{P(A)}=0.6$.

We know $P(A)=0.4$.

So , $\dfrac{P(A\cap B)}{0.4}=0.6$.

$\Rightarrow P(A\cap B)=0.6\times 0.4=0.24$.

Now , we have to find the probability of happening of the event $A$ given that event $B$ has already happened , i.e. we have to find the value of $P\left( \dfrac{A}{B} \right)$.

We know , $P\left( \dfrac{A}{B} \right)=\dfrac{P(A\cap B)}{P(B)}$ .

So , $P\left( \dfrac{A}{B} \right)=\dfrac{0.24}{0.8}=0.3$.

Hence , the value of $P\left( \dfrac{A}{B} \right)$ is equal to $0.3$.

Note: Students generally get confused between $P\left( \dfrac{A}{B} \right)$ and $P\left( \dfrac{B}{A} \right)$. Both are not the same . $P\left( \dfrac{A}{B} \right)$ gives the probability of happening of event $A$ given that event $B$ has already happened , whereas $P\left( \dfrac{B}{A} \right)$ gives the probability of happening of event $B$ given that event $A$ has already happened.

Before we proceed with the solution , we must know the concept of conditional probability.

When the probability of happening of one event is affected by the happening of another, then the two events are known as dependent events.

For example : There are $2$ black cards and $5$ red cards in a pile. The probability of picking a red card is given as $\dfrac{5}{7}$ . But , after removing the red card , now the probability of picking a red or a black card changes . The probability of picking a red card becomes $\dfrac{4}{6}=\dfrac{2}{3}$ and the probability of picking a black card becomes $\dfrac{2}{6}=\dfrac{1}{3}$ .

Now , if $A$ and \[B\] are two dependent events , then the probability of happening of $A$ given that \[B\] has already happened is given by $P\left( \dfrac{A}{B} \right)=\dfrac{P(A\cap B)}{P(B)}$ , where $P(B)$ is the probability of happening of event \[B\] and $P(A\cap B)$ is the probability of happening of events $A$ and \[B\] simultaneously.

Now , coming to the question , we are given probability of happening of event $A$ is equal to $0.4$ , the probability of happening of event $B$ is equal to $0.8$ and the probability of happening of the event $B$ given that $A$ has happened is $0.6$ .

Now ,from the information that the probability of happening of the event $B$ given that $A$ has happened is $0.6$, we can write $\dfrac{P(A\cap B)}{P(A)}=0.6$.

We know $P(A)=0.4$.

So , $\dfrac{P(A\cap B)}{0.4}=0.6$.

$\Rightarrow P(A\cap B)=0.6\times 0.4=0.24$.

Now , we have to find the probability of happening of the event $A$ given that event $B$ has already happened , i.e. we have to find the value of $P\left( \dfrac{A}{B} \right)$.

We know , $P\left( \dfrac{A}{B} \right)=\dfrac{P(A\cap B)}{P(B)}$ .

So , $P\left( \dfrac{A}{B} \right)=\dfrac{0.24}{0.8}=0.3$.

Hence , the value of $P\left( \dfrac{A}{B} \right)$ is equal to $0.3$.

Note: Students generally get confused between $P\left( \dfrac{A}{B} \right)$ and $P\left( \dfrac{B}{A} \right)$. Both are not the same . $P\left( \dfrac{A}{B} \right)$ gives the probability of happening of event $A$ given that event $B$ has already happened , whereas $P\left( \dfrac{B}{A} \right)$ gives the probability of happening of event $B$ given that event $A$ has already happened.

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