Question

If $A$ and $B$ are two independent events such that $P\left(\bar{A}\cap B\right)=\frac{2}{15}$ and $P\left(A\cap \bar{B}\right)=\frac{1}{6}$, then find $P(A)$ and $P(B)$ respectively.

• A. $\frac{5}{6}$, $\frac{4}{5}$
• B. $\frac{1}{5}$, $\frac{1}{7}$
• C. $\frac{1}{6}$, $\frac{1}{7}$
• D. $\frac{1}{7}$, $\frac{1}{7}$

Answer 1

Answer: (A)

Since $A$ and $B$ are independent events, therefore, $A^c$ and $B$ are independent and also $A$ and $B^c$ are independent.
$\therefore P\left(\bar{A}\cap B\right)=P\left(A^c\cap B\right)=P\left(A^c\right)P\left(B\right)$
$=\left(1-P\left(A\right)\right)P\left(B\right)$
and $P\left(A\cap \bar{B}\right)=P\left(A\cap B^c\right)=P\left(A\right)P\left(B^c\right)$
$=P\left(A\right)\left(1-P\left(B\right)\right)$
$\left(\because \bar{A}=A^c\text{and}\bar{B}=B^c\right)$
$\Rightarrow \left(1-P\left(A\right)\right)P\left(B\right)=P\left(\bar{A}\cap B\right)=\frac{2}{15}\dots \left(1\right)$
and $P\left(A\right)\left(1-P\left(B\right)\right)=P\left(A\cap \bar{B}\right)=\frac{1}{6}\dots \left(2\right)$
Subtracting $\left(2\right)$ from $\left(1\right)$, we obtain
$P\left(B\right)-P\left(A\right)=\frac{2}{15}-\frac{1}{6}$
$\Rightarrow P\left(B\right)-P\left(A\right)=\frac{4-5}{30}$
$\Rightarrow P\left(B\right)-P\left(A\right)=-\frac{1}{30}\dots \left(3\right)$
Substituting this value of $P\left(B\right)$ in $\left(1\right)$, we get
$\left(1-P\left(A\right)\right)\left\{P\left(A\right)-\frac{1}{30}\right\}=\frac{2}{15}$
$\Rightarrow P\left(A\right)+\frac{1}{30}P\left(A\right)-{\left(P\left(A\right)\right)}^2-\frac{1}{30}=\frac{2}{15}$
$\Rightarrow {\left(P\left(A\right)\right)}^2-\frac{31}{30}P\left(A\right)+\frac{1}{30}+\frac{2}{15}=0$
$\Rightarrow 30{\left(P\left(A\right)\right)}^2-31P\left(A\right)+5=0$
$\Rightarrow P\left(A\right)=\frac{31\pm \sqrt{961-600}}{60}$
$=\frac{31\pm 19}{60}=\frac{5}{6},\frac{1}{5}$
Case I. When $P\left(A\right)=\frac{5}{6}$, then from $\left(3\right)$,
$P\left(B\right)=P\left(A\right)-\frac{1}{30}=\frac{5}{6}-\frac{1}{30}=\frac{25-1}{30}=\frac{4}{5}$
$\Rightarrow P\left(A\right)=\frac{5}{6}$ and $P\left(B\right)=\frac{4}{5}$
Case II. When $P\left(A\right)=\frac{1}{5}$, then from $\left(3\right)$,
$P\left(B\right)=P\left(A\right)-\frac{1}{30}=\frac{1}{5}-\frac{1}{30}=\frac{6-1}{30}=\frac{1}{6}$
$\Rightarrow P\left(A\right)=\frac{1}{5}$ and $P\left(B\right)=\frac{1}{6}$.