Question

If $A$ and $B$ are independent events such that $P(B)=\frac{2}{7},P\left(A\cup \bar{B}\right)=0.8,thenP(A)$

• A. 0.4
• B. 0.3
• C. 0.2
• D. 0.1

Answer 1

Answer: (B)

We know that,
$P(A\cup \bar{B})=P(A)+P(\bar{B})-P(A\cap \bar{B})$
$=P(A)+[1-P(B)]-P(A)\cdot P(\bar{B})$
$=P(A)+[1-P(B)]-P(A)[1-P(B)]$
$\Rightarrow 0.8=P(A)+(1-\frac{2}{7})-P(A)(1-\frac{2}{7})$ [Given]
$\Rightarrow \frac{4}{5}=P(A)[1-1+\frac{2}{7}]+\frac{5}{7}$
$\Rightarrow \frac{4}{5}-\frac{5}{7}=P(A)\cdot \frac{2}{7}$
$\Rightarrow \frac{28-25}{35}=P(A)\cdot \frac{2}{7}$
$\Rightarrow P(A)=\frac{3}{35}\cdot \frac{7}{2}$
$=\frac{3}{5\times 2}$
$=0.3$