Question

The probability of a shooter hitting a target is 3/4 How many minimum number of times must he fire so that the probability of hitting the target at least once is more than 90%?

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is B

The probability of hitting the target at least once is the complement of the probability of missing the target every time. Let the number of shots fired be n.

The probability of missing the target in a single shot is:

\[ P(\text{miss}) = 1 - \frac{3}{4} = \frac{1}{4}. \]

The probability of missing the target in all n shots is:

\[ P(\text{miss all}) = \left( \frac{1}{4} \right)^n. \]

The probability of hitting the target at least once is:

\[ P(\text{hit at least once}) = 1 - P(\text{miss all}) = 1 - \left( \frac{1}{4} \right)^n. \]

We need \(P(\text{hit at least once})>0.9 :\)

\(1 - \left( \frac{1}{4} \right)^n >0.9.\)

Simplify:

\(\left( \frac{1}{4} \right)^n<0.1.\)

Take the logarithm (base 10) of both sides:

\(n \log_{10} \left( \frac{1}{4} \right)<\log_{10}(0.1).\)

Using \( \log_{10} \left( \frac{1}{4} \right) = -\log_{10}(4) \) and \( \log_{10}(0.1) = -1 \):

\(-n \log_{10}(4)<-1.\)

Divide by \( -\log_{10}(4) \) (note the sign change):

\(n>\frac{1}{\log_{10}(4)}.\)

Approximate \( \log_{10}(4) \approx 0.602 \):

\(n>\frac{1}{0.602} \approx 1.66.\)

Since n must be an integer, the minimum n is:

\[ n = 2. \]

Thus, the shooter must fire at least 2 times.