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) \[\therefore \,\,\,{{\left( \frac{9}{10} \right)}^{n}}=\frac{1}{2}\], \[\therefore \,\,\,n\left\{ 2{{\log }_{10}}3-1 \right\}=-{{\log }_{10}}2\] \[\therefore \,\,\,n=\left\{ \frac{{{\log }_{10}}2}{1-12{{\log }_{10}}3}=\frac{0.3010}{1-2\times 0.4771}=6.5 \right.\] (nearly) \[\therefore \] For 6 shots, the probabilty is about 53% while for 7 shots it is nearly 48%.
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