Probability Question 302
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 target, at least once is more than 0.9, is
[use $ \log 4=0.602and\log 3=0.477 $ ]
Options:
A) 7
B) 8
C) 6
D) 5
Show Answer
Answer:
Correct Answer: B
Solution:
Let n denote the required number of shots and X the number of shots that hit the target. Then $ X\tilde{\ }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{{( \frac{3}{4} )}^{n}}\ge 0.9\Rightarrow {{( \frac{3}{4} )}^{n}}\le \frac{1}{10} $
$ \Rightarrow {{( \frac{4}{3} )}^{n}}\ge 10\Rightarrow n(log4-log3)\ge 1 $
$ \Rightarrow n(0.602-0.477)\ge 1\Rightarrow n\ge \frac{1}{0.125}=8 $ Therefore the least number of trials required is 8.
BETA
