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

Hint:

When dealing with probabilities involving repeated independent events, it’s often helpful to use the complement rule. Instead of calculating the probability of success in every shot, calculate the probability of failure in each shot and use the complement rule to find the overall probability. Also, when dealing with inequalities involving exponents, logarithms are a powerful tool for simplifying the problem and solving for unknowns like the number of shots required. Always remember to approximate logarithms and reverse the inequality when dividing by negative numbers!

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.

Answer 2

The probability of hitting the target at least once is the complement of the probability of missing the target in every shot. Let the number of shots fired be denoted by \(n\).

Step 1: Probability of missing the target in a single shot:

Given that the probability of hitting the target in one shot is \( \frac{3}{4} \), the probability of missing the target on any given shot is the complement of this probability, which is:

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

Step 2: Probability of missing the target in all \(n\) shots:

Since the shots are independent, the probability of missing the target in every one of the \(n\) shots is:

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

Step 3: Probability of hitting the target at least once:

The probability of hitting the target at least once is the complement of the probability of missing the target in all shots:

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

Step 4: Solve the inequality to find the minimum number of shots:

We want the probability of hitting the target at least once to be greater than 0.9:

\[ P(\text{hit at least once}) > 0.9. \]

Substituting the expression for \( P(\text{hit at least once}) \), we get:

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

Simplifying the inequality:

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

Step 5: Take the logarithm of both sides:

To solve for \(n\), we apply the logarithm (base 10) to 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 \( \log_{10}(0.1) = -1 \), we simplify to:

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

Step 6: Solve for \(n\):

Dividing both sides by \( -\log_{10}(4) \) reverses the inequality:

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

Approximating \( \log_{10}(4) \approx 0.602 \), we find:

\[ n > \frac{1}{0.602} \approx 1.66. \]

Step 7: Find the minimum integer value for \(n\):

Since \(n\) must be an integer, we round up to the next whole number:

\[ n = 2. \]

Conclusion: Therefore, the shooter must fire at least 2 times to have a greater than 0.9 chance of hitting the target at least once.