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\left(\log4 - \log3\right)\ge1$
$ \Rightarrow n\left(0.602 - 0.477\right) \ge1 $
$\Rightarrow n \ge \frac{1}{0.125} = 8 $
Therefore the least number of trials required is 8.