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

Here, $P(A) = 0.4,P(\bar A) = 0.6,P(B) = 0.3,P(\bar B) = 0.7$,
$P(C) = 0.2$ and $P(\bar C) = 0.8$

$\therefore$ Probability of two hits $= {P_A} \cdot {P_B} \cdot {P_{\bar C}} + {P_A} \cdot {P_{\bar B}} \cdot {P_C} + {P_{\bar A}} \cdot {P_B} \cdot {P_C}$

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

$= 0.096 + 0.056 + 0.036 = 0.188$