Question

If $A$ and $B$ are two independent events such that $P\left(A\right)>\frac{1}{2}, \, P\left(A \cap B^{C}\right)=\frac{3}{25}$ and $P\left(A^{C} \cap B\right)=\frac{8}{25},$ then $P\left(A\right)$ is equal to (where, $A^{c}$ and $B^{c}$ represent the complement of events $A$ and $B$ respectively)

• A. $\frac{1}{5}$
• B. $\frac{3}{5}$
• C. $\frac{3}{4}$
• D. $\frac{4}{5}$

Answer 1

Answer: (B)

$A, B$ are independent events $P\left(A \cap B^{C}\right)=P(A) P\left(B^{C}\right)=P(A)(1-P(B))=\frac{3}{25} \ldots$
$P\left(A^{C} \cap B\right)=P\left(A^{C}\right) \cdot P(B)=(1-P(A)) P(B)=\frac{8}{25} \ldots(2)$
From (1) and (2), we get, $P(B)-P(A)=\frac{1}{5} \ldots$ (3)
From (1) and (3) $\left(P(B)-\frac{1}{5}\right)(1-P(B))=\frac{3}{25}$
Let, $P(B)=t$ $\Rightarrow \frac{6}{5} t-\frac{1}{5}-t^{2}=\frac{3}{25} \Rightarrow 30 t-5-25 t^{2}=3$
$\Rightarrow 25 t^{2}-30 t+8=0 \Rightarrow t=\frac{2}{5}, \frac{4}{5}$
$P(B)=\frac{2}{5} \Rightarrow P(A)=\frac{1}{5}$
$P(B)=\frac{4}{5} \Rightarrow P(A)=\frac{3}{5}$
$\because P(A)>\frac{1}{2} \Rightarrow P(A)=\frac{3}{5}$