Probability 5 Question 6
6. If the probability of hitting a target by a shooter in any shot, is $\frac{1}{3}$, then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than $\frac{5}{6}$, is
(2019 Main, 10 Jan II)
(a) 6
(b) 3
(c) 5
(d) 4
Show Answer
Answer:
Correct Answer: 6. (c)
Solution:
- The probability of hitting a target at least once $=1-($ probability of not hitting the target in any trial) $=1-{ }^{n} C _0 p^{0} q^{n}$
where $n$ is the number of independent trials and $p$ and $q$ are the probability of success and failure respectively.
[by using binomial distribution]
Here, $\quad p=\frac{1}{3}$ and $\quad q=1-p=1-\frac{1}{3}=\frac{2}{3}$
According to the question, $1-{ }^{n} C _0 \quad \frac{1}{3}{ }^{0} \quad \frac{2}{3}{ }^{n}>\frac{5}{6}$
$\Rightarrow \frac{2}{3}^{n}<1-\frac{5}{6} \Rightarrow \frac{2}{3}^{n}<\frac{1}{6}$
Clearly, minimum value of $n$ is 5 .
