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  • 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 A. 0.024 B. 0.188 C. 0.336 D. 0.452

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
A. 0.024
B. 0.188
C. 0.336
D. 0.452

Answers (1)

Given-
\\\mathrm{P}(\mathrm{A})=0.4 \mathrm{P}(\mathrm{B})=0.3$ and $\mathrm{P}(\mathrm{C})=0.2$ \\Therefore, $\mathrm{P}\left(\mathrm{A}^{\prime}\right)=1-\mathrm{P}(\mathrm{A})=[1-0.4]=0.6$ \\$\mathrm{P}\left(\mathrm{B}^{\prime}\right)=1-\mathrm{P}(\mathrm{B})=[1-0.3]=0.7$ \\$\mathrm{P}\left(\mathrm{C}^{\prime}\right)=1-\mathrm{P}(\mathrm{C})=[1-0.2]=0.8$ \\$\mathrm{P}(\mathrm{E})=\left[\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}\left(\mathrm{C}^{\prime}\right)\right]+\left[\mathrm{P}(\mathrm{A}) \times \mathrm{P}\left(\mathrm{B}^{\prime}\right) \times \mathrm{P}(\mathrm{C})\right]+\left[\mathrm{P}\left(\mathrm{A}^{\prime}\right) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}(\mathrm{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)] \\ =0.96+0.056+0.036 \\$=0.188$
Hence, Probability of two hits is 0.188

Option B is correct.

Posted by

infoexpert22

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