Question

The probability of a man hitting a target is $1 / 4$. The number of times he must shoot so that the probability he hits the target, at least once is more than $0.9$, is
[use $\log 4=0.602$ and $\log 3=0.477]$

• A. 7
• B. 8
• C. 6
• D. 5

Answer 1

Answer: (B)

Let $n$ denote the required number of shots and $X$ the number of shots that hit the target. Then $X \sim B ( n , p )$
with $p =1 / 4$. Now, $P ( X \geq 1) \geq 0.9$
$ \Rightarrow 1- P ( X =0) \geq 0.9$
$\Rightarrow 1-{ }^{n} C_{0}\left(\frac{3}{4}\right)^{n} \geq 0.9$
$ \Rightarrow \left(\frac{3}{4}\right)^{n} \leq \frac{1}{10}$
$\Rightarrow \left(\frac{4}{3}\right)^{ n } \geq 10 $
$\Rightarrow n (\log \,4-\log 3) \simeq 1$
$\Rightarrow n (0.602-0.477) \geq 1$
$ \Rightarrow n \geq \frac{1}{0.125}=8$