Question

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:
1. 3
2. 6
3. 5
4. 4

Answer 1

Correct Answer - Option 3 : 5

The probability of hitting a target at least once is given below:

\(P = 1\) - (probability of not hitting the target in any trial)

By using binomial distribution,

⇒ P = 1 - ⁿC₀ p⁰ qⁿ

Where

‘n’ is the number of independent trials.

‘p’ and ‘q’ are the probability of success and failure respectively.

From question,

\(\Rightarrow p = \frac{1}{3}\)

∵ [p + q = 1 ⇒ q = 1 - p]

\(\Rightarrow q = 1 - \frac{1}{3}\)

\(\therefore q = \frac{2}{3}\)

From question,

\(\Rightarrow 1{ - ^n}{C_0}{\left( {\frac{1}{3}} \right)^0}{\left( {\frac{2}{3}} \right)^n} > \frac{5}{6}\)

\(\Rightarrow {\left( {\frac{2}{3}} \right)^n} < 1 - \frac{5}{6}\)

\(\Rightarrow {\left( {\frac{2}{3}} \right)^n} < \frac{1}{6}\)

Thus, the minimum value of n is 5.