Question

If $A$ and $B$ are two independent events such that$P(A\prime \cap B)=\frac{2}{15}$ and$P(A\cap B\prime )=\frac{1}{6}$, the $P\left(B\right)$ is

Answer 1

Step 1: Use properties of independent events to generate equations.

Given,

$P(A\prime \cap B)=\frac{2}{15}$

$P(A\cap B\prime )=\frac{1}{6}$

It is given that $A$ and $B$ and are independent. Thus,

$\begin{array}{rcl}\mathit{P}\mathbf{(}\mathit{A}\mathbf{\text{'}}\mathbf{\cap }\mathit{B}\mathbf{)}& \mathbf{=}& \mathit{P}\mathbf{(}\mathit{A}\mathbf{\text{'}}\mathbf{)}\mathbf{\times }\mathit{P}\mathbf{\left(}\mathit{B}\mathbf{\right)}\\ & =& [1-P(A\left)\right]\times P\left(B\right)=\frac{2}{15}\dots \left(\mathrm{i}\right)\end{array}$.

$\begin{array}{rcl}\mathit{P}\mathbf{(}\mathit{A}\mathbf{\cap }\mathit{B}\mathbf{\text{'}}\mathbf{)}& \mathbf{=}& \mathit{P}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{\times }\mathit{P}\mathbf{(}\mathit{B}\mathbf{\text{'}}\mathbf{)}\\ & =& P\left(A\right)\times [1-P(B\left)\right]=\frac{1}{6}\dots \left(\mathrm{ii}\right)\end{array}$

Upon subtracting $\left(\mathrm{i}\right)$ from $\left(\mathrm{ii}\right)$ we get,

$P\left(A\right)-P\left(B\right)=\frac{1}{6}-\frac{2}{15}$

$\Rightarrow P\left(A\right)-P\left(B\right)=\frac{1}{30}$

$\Rightarrow$ $P\left(A\right)=\frac{1}{30}+P\left(B\right)$. $\dots \left(\mathrm{iii}\right)$

Step 2: Substitute equation $\left(\mathrm{iii}\right)$ in equation $\left(\mathrm{ii}\right)$ and solve the resulting quadratic equation.

Substituting $\left(\mathrm{iii}\right)$ in $\left(\mathrm{ii}\right)$we get,

$\left[\frac{1}{30}+P\left(B\right)\right]·\left[1-P\left(B\right)\right]=\frac{1}{6}$

$\Rightarrow$$-P{\left(B\right)}^2-\frac{1}{30}P\left(B\right)+P\left(B\right)+\frac{1}{30}=\frac{1}{6}$

$\Rightarrow$ $P{\left(B\right)}^2-\frac{29}{30}P\left(B\right)+\frac{4}{30}=0$

$\Rightarrow$ $30P{\left(B\right)}^2-29P\left(B\right)+4=0$

This is a quadratic equation in $P\left(B\right)$.

For a quadratic equation of the form $a{x}^2+bx+c=0$ where $a\ne 0$, the roots of theequation are given as,

$\mathit{x}\mathbf{=}\frac{\mathbf{-}\mathbf{b}\mathbf{\pm }\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}$

Using the above formula,

$P\left(B\right)=\frac{-(-29)\pm \sqrt{{(-29)}^2-4\times 30\times 4}}{2\times 30}$

$$\begin{gathered} \Rightarrow P\left(B\right)=\frac{29\pm \sqrt{361}}{60} \\ \Rightarrow P\left(B\right)=\frac{29\pm 19}{60} \end{gathered}$$

$\Rightarrow P\left(B\right)=\frac{4}{5}$ or $P\left(B\right)=\frac{1}{6}$

Hence the value of $P\left(B\right)$ is $\frac{4}{5}$ or $\frac{1}{6}$.