Question

Question: Suppose thatandare events in a sample space and$\mathit{p}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{,}\mathit{p}\mathbf{\left(}\mathit{F}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$, and $\mathit{p}\mathbf{(}\mathit{E}\mathbf{\mid }\mathit{F}\mathbf{)}\mathbf{=}\frac{\mathbf{2}}{\mathbf{5}}$. Find$\mathit{p}\mathbf{(}\mathit{F}\mathbf{\mid }\mathit{E}\mathbf{)}$.

Short answer

Answer:

The value of$p\left(F\mid E\right)$ is$\frac{3}{5}$ .

Step-by-step solution

Given data

In a sample space,$E$ and $F$are the events.

$p\left(E\right)=\frac{1}{3},p\left(F\right)=\frac{1}{2}$, and$p\left(E\mid F\right)=\frac{2}{5}$

The value of $p\left(F\mid E\right)$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, $p\left(B\right)$is the probability of the second event, and $p\left(A\mid B\right)$is the conditional probability.

Find the conditional probability

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

$$\begin{gathered} \Rightarrow \frac{2}{5}=\frac{p(E\cap F)}{\frac{1}{2}} \\ \Rightarrow p(E\cap F)=\frac{2}{5}\times \frac{1}{2} \\ \Rightarrow p(E\cap F)=\frac{1}{5} \end{gathered}$$

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

$$\begin{gathered} \Rightarrow p\left(F\mid E\right)=\frac{\frac{1}{5}}{\frac{1}{3}} \\ \Rightarrow p\left(F\mid E\right)=\frac{1}{5}\times \frac{3}{1} \\ \Rightarrow p\left(F\mid E\right)=\frac{3}{5} \end{gathered}$$

Hence, the value of$p\left(F\mid E\right)$ is $\frac{3}{5}$.