A) 7
B) 8
C) 9
D) 6
Correct Answer: A
Solution :
| The probability of hitting in one shot \[=\frac{10}{100}=\frac{1}{10}\] |
| If he fires n shots, the probability of hitting at least once |
| \[=1-{{\left( 1-\frac{1}{10} \right)}^{n}}=1-{{\left( \frac{9}{10} \right)}^{n}}=\frac{1}{2}\] (from the question) |
| \[\therefore \,\,{{\left( \frac{9}{10} \right)}^{n}}=\frac{1}{2},\] \[\therefore \,\,n\{2{{\log }_{10}}3-1\}=-\log {{ & }_{10}}2\] |
| \[\therefore \,\,\,n=\frac{{{\log }_{10}}2}{1-2{{\log }_{10}}3}=\frac{0.3010}{1-2\times 0.4771}=6.5\] (nearly) |
| \[\therefore \] for 6 shots, the probability is about \[47%\] while for 7 shots it is nearly\[53%\]. |
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