Question

Let $A$ and $B$ be two independent events such that their probabilities are $\frac{3}{10}$ and $\frac{2}{5}$. The probability of exactly one of the events happening is

• A. $\frac{23}{50}$
• B. $\frac{1}{2}$
• C. $\frac{31}{50}$
• D. none of these

Answer 1

Answer: (A)

$\frac{23}{50}$

Let E be the desired event.
so $P\left(E\right)=P(A\cap \overline{B})+P(B\cap \overline{A})=P\left(A\right)-P(A\cap B)+P\left(B\right)-P(A\cap B)=P\left(A\right)-2P(A\cap B)+P\left(B\right)=P\left(A\right)-2\left(P\left(A\right)P\left(B\right)\right)+P\left(B\right)$
since A and B events are independent.
$\Rightarrow P\left(E\right)=\frac{3}{10}-2(\frac{3}{10}\times \frac{2}{5})+\frac{2}{5}\Rightarrow P\left(E\right)=\frac{15-12+20}{50}=\frac{23}{50}$