Question

If $A$ and $B$ are independent events associated to some experiment $E$ such that $P\left(A^c\cap B\right)=\frac{2}{15}$ and $P\left(A\cap B^c\right)=\frac{1}{6}$, then $P\left(B\right)$ , then $P(B)$ is equal to

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

Answer 1

Answer: (B)

We have, $P(A^c\cap B)=\frac{2}{15}$ and $P(A\cap B^c)=\frac{1}{6}$
$\Rightarrow P(A^c)\cdot P(B)=\frac{2}{15}$ and $P(A)\cdot P(B^c)=\frac{1}{6}$
$\Rightarrow [1-P(A)\cdot P(B)=\frac{2}{15}$ and $P(A)\cdot [1-P(B)]=\frac{1}{6}$
$\Rightarrow [1-P(A)]\cdot P(B)=\frac{2}{15}$ and $P(A)=\frac{1}{6[1-P(B)]}$
$\therefore [1-\frac{1}{6[1-P(B)]}]\cdot P(B)=\frac{2}{15}$
$\Rightarrow \frac{[6-6P(B)-1]}{6[1-P(B)]}\cdot P(B)=\frac{2}{15}$
$\Rightarrow 15[5-6P(B)]P(B)=2\times 6[1-P(B)]$
$\Rightarrow 75P(B)-90P(B{)}^2=12-12P(B)$
Let $P(B)=x$
$\therefore 75x-90x^2=12-12x$
$\Rightarrow 90x^2-75x-12x+12=0$
$\Rightarrow 90x^2-87x+12=0$
$\Rightarrow 90x^2-15x-72x+12=0$
$\Rightarrow 15x(6x-1)-12(6x-1)=0$
$\Rightarrow (6x-1)(15x-12)=0$
$\Rightarrow 6x-1=0$ or $15x-12=0$
$\Rightarrow x=\frac{1}{6}$ or $x=\frac{12}{15}=0$
$\Rightarrow x=\frac{1}{6}$ or $x=\frac{4}{5}$