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Question
Mark the correct alternative in the following question
Three persons, A, B and C fire a target in turn starting with A. Their probabilities of hitting the target are 0.4, 0.2 and 0.2, respectively. The probability of two hits is
Options
0.024
0.452
0.336
0.188
Solution
\[\text{ Let } : \]
\[\text{ A be the event of hitting the target by the person A } , \]
\[\text{ B be the event of hitting the target by the person B and } \]
\[\text{ C be the event of hitting the target by the person C } \]
\[\text{ We have} , \]
\[P\left( A \right) = 0 . 4, P\left( B \right) = 0 . 3 \text{ and } P\left( C \right) = 0 . 2\]
\[\text{ Also} , \]
\[P\left(\overline{A} \right) = 1 - P\left( A \right) = 1 - 0 . 4 = 0 . 6, \]
\[P\left( \overline{B} \right) = 1 - 0 . 3 = 0 . 7 \text{ and } \]
\[P\left( \overline{C} \right) = 1 - 0 . 2 = 0 . 8\]
\[\text{ Now } , \]
\[P\left( \text{ Two hits } \right) = P\left( AB\overline{C} \right) + P\left( A\overline{B}C \right) + P\left( \overline{A}BC \right)\]
\[ = P\left( A \right) \times P\left( B \right) \times P\left(\overline{ C } \right) + P\left( A \right) \times P\left( \overline{B} \right) \times P\left( C \right) + P\left(\overline{ A} \right) \times P\left( B \right) \times P\left( C \right)\]
\[ = 0 . 4 \times 0 . 3 \times 0 . 8 + 0 . 4 \times 0 . 7 \times 0 . 2 + 0 . 6 \times 0 . 3 \times 0 . 2\]
\[ = 0 . 096 + 0 . 056 + 0 . 036\]
\[ = 0 . 188\]
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