Question

Question: Suppose that $\mathit{E}$and$\mathit{F}$are events in a sample space and$\mathit{p}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{,}\mathit{p}\mathbf{\left(}\mathit{F}\mathbf{\right)}\mathbf{=}\frac{\mathbf{3}}{\mathbf{4}}\mathbf{,}$and $\mathit{p}\mathbf{(}\mathit{F}\mathbf{\mid }\mathit{E}\mathbf{)}\mathbf{=}\frac{\mathbf{5}}{\mathbf{8}}$. Find$\mathit{p}\mathbf{(}\mathit{E}\mathbf{\mid }\mathit{F}\mathbf{)}$.

Short answer

Answer:

The value of $P\left(E|F\right)$is $\frac{5}{9}$.

Step-by-step solution

Given data

In a sample space, and are the events.

$p\left(E\right)=\frac{2}{3},p\left(F\right)=\frac{3}{4}$, and$\text{p}(F\mid E)=\frac{5}{8}$

The value of$P\left(E|F\right)$ is to be found.

Definition and formula to be used

The likelihood of an event or outcome occurring owing to the existence of a preceding event or outcome is defined as conditional probability.

The formula for conditional probability is,

$p\left(A\mid B\right)=\frac{p(A\cap B)}{p\left(B\right)}$

Here, $A,B$are two events, $p(A\cap B)$is the probability of intersection of the two events, is the probability of the second event, $p\left(B\right)$and$p\left(A\mid B\right)$ is the conditional probability.

Find the conditional probability

In the formula $p\left(F\mid E\right)=\frac{p(E\cap F)}{p\left(E\right)}$, substitute the given values.

$$\begin{gathered} \Rightarrow \frac{5}{8}=\frac{p(E\cap F)}{\frac{2}{3}} \\ \Rightarrow p(E\cap F)=\frac{5}{8}\times \frac{2}{3} \\ \Rightarrow p(E\cap F)=\frac{5}{12} \end{gathered}$$

Now, substitute the values in the formula $p\left(E\mid F\right)=\frac{p(E\cap F)}{p\left(F\right)}$.

$$\begin{gathered} \Rightarrow p\left(E\mid F\right)=\frac{\frac{5}{12}}{\frac{3}{4}} \\ \Rightarrow p\left(E\mid F\right)=\frac{5}{12}\times \frac{4}{3} \\ \Rightarrow p\left(E\mid F\right)=\frac{5}{9} \end{gathered}$$

Hence, the value of $p\left(E\mid F\right)$is $\frac{5}{9}$.