Question

The probability of a shooter hitting a target is 2/4 . How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more 0.99?

Answer 1

Answer on Question #63773 – Math – Statistics and Probability

Question

The probability of a shooter hitting a target is $2/4$. How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more 0.99?

Solution

Let $p$ denote the probability of a shooter hitting a target at each of $n$ times. Then by the product rule for $n$ independent events the probability of hitting the target $n$ times is $p^n$ while the probability of not hitting the target is $(1 - p)^n$.

Finally, the probability of hitting the target at least once is $1 - (1 - p)^n$.

Therefore, we have

$$\begin{array}{l} 1 - (1 - 1/2)^n > 0.99; \\ 1 - (1/2)^n > 0.99; \\ 1 - 0.99 > (1/2)^n; \\ 1/2^n < 0.01; \\ \frac{1}{0.01} < 2^n; \\ 2^n > 100; \\ \log 2^n > \log 10^2; \\ n \cdot \log 2 > 2 \log 10; \\ n > \frac{2}{\log 2}; \\ n > 6.64. \end{array}$$

Thus, the minimum number of times is $n = 7$.

Check:

$$1 - \frac{1}{2^6} \approx 0.984; \quad 1 - \frac{1}{2^7} \approx 0.992.$$

Answer: 7.

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