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, $P(X\ge 1)\ge 0.9$
$\Rightarrow 1-P(X=0)\ge 0.9$
$\Rightarrow 1-{}^n{C}_0{\left(\frac{3}{4}\right)}^n\ge 0.9$
$\Rightarrow {\left(\frac{3}{4}\right)}^n\le \frac{1}{10}$
$\Rightarrow {\left(\frac{4}{3}\right)}^n\ge 10$
$\Rightarrow n(\log 4-\log 3)\simeq 1$
$\Rightarrow n(0.602-0.477)\ge 1$
$\Rightarrow n\ge \frac{1}{0.125}=8$