Three persons $A$, $B$ and $C$, fire at a target in turn, starting with $A$. Their probability of hitting the target are $0.4$, $0.3$ and $0.2$, respectively. The probability of two hits is

$0.188$

The correct answer is Option (2) → $0.188$ ##

Here, $P(A) = 0.4$

$\Rightarrow P(\bar{A}) = 0.6, P(B) = 0.3 \Rightarrow P(\bar{B}) = 0.7$ and $P(C) = 0.2 \Rightarrow P(\bar{C}) = 0.8$

$∴\text{Probability of two hits}$

$= P(A) \cdot P(B) \cdot P(\bar{C}) + P(A) \cdot P(\bar{B}) \cdot P(C) + P(\bar{A}) \cdot P(B) \cdot P(C)$

$= 0.4 \times 0.3 \times 0.8 + 0.4 \times 0.7 \times 0.2 + 0.6 \times 0.3 \times 0.2$

$= 0.096 + 0.056 + 0.036 = 0.188$