Choose the correct answer from the given four options.Three perso… - Vidyadip

MCQ

Choose the correct answer from the given four options.Three persons$, A, B$ and $C,$ fire at a target in turn, starting with $A.$ Their probabilityof hitting the target are $0.4, 0.3$ and $0.2$ respectively. The probability of two hits is:

A
$0.024$
✓
$0.188$
C
$0.336$
D
$0.452$

✓

Answer

Correct option: B.

$0.188$

We have
$\text{P}(\text{A})=0.4,\text{P}(\bar{\text{A}})=0.6,\text{P}(\text{B})=0.3,\text{P}(\bar{\text{B}})=0.7$
$\text{P}(\text{C})=0.2$ and $\text{P}(\bar{\text{C}})=0.8$
$\therefore$ Probability of two hits $=\text{P}_{\text{A}}\cdot\text{P}_{\text{B}}\cdot\text{P}_{\bar{\text{C}}}+\text{P}_{\text{A}}\cdot\text{P}{_\bar{\text{B}}}\cdot\text{P}_{\text{C}}+\text{P}{_\bar{\text{A}}}\cdot\text{P}_{\text{B}}\cdot\text{P}_{\text{C}}$
$=0.4\times0.3\times0.8+0.4\times0.7\times0.2+0.6\times0.3\times0.2$
$=0.096+0.056+0.036=0.188$

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