Question
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 probability
of hitting the target are 0.4, 0.3 and 0.2 respectively. The probability of two hits
is:
  1. 0.024
  2. 0.188
  3. 0.336
  4. 0.452
 

Answer

  1. 0.188
Solution:
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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