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

Answer 1

\[\text{ Let } : \]

\[\text{ A be the event of hitting the target by the person A } , \]

\[\text{ B be the event of hitting the target by the person B and } \]

\[\text{ C be the event of hitting the target by the person C } \]

\[\text{ We have} , \]

\[P\left( A \right) = 0 . 4, P\left( B \right) = 0 . 3 \text{ and }  P\left( C \right) = 0 . 2\]

\[\text{ Also} , \]

\[P\left(\overline{A} \right) = 1 - P\left( A \right) = 1 - 0 . 4 = 0 . 6, \]

\[P\left( \overline{B} \right) = 1 - 0 . 3 = 0 . 7 \text{ and } \]

\[P\left( \overline{C} \right) = 1 - 0 . 2 = 0 . 8\]

\[\text{ Now } , \]

\[P\left( \text{ Two hits } \right) = P\left( AB\overline{C} \right) + P\left( A\overline{B}C \right) + P\left( \overline{A}BC \right)\]

\[ = P\left( A \right) \times P\left( B \right) \times P\left(\overline{ C } \right) + P\left( A \right) \times P\left( \overline{B} \right) \times P\left( C \right) + P\left(\overline{ A} \right) \times P\left( B \right) \times P\left( C \right)\]

\[ = 0 . 4 \times 0 . 3 \times 0 . 8 + 0 . 4 \times 0 . 7 \times 0 . 2 + 0 . 6 \times 0 . 3 \times 0 . 2\]

\[ = 0 . 096 + 0 . 056 + 0 . 036\]

\[ = 0 . 188\]