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:
- A0.024
- B0.188
- C0.336
- D0.452
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:
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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| X = xi | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X = Xi) | 0 | 2p | 2p | 3p | p2 | 2p2 | 7p2 | 2p |