Step 1: Understanding the Question:
This problem describes a scenario with a fixed number of independent trials (5 shots), each with two possible outcomes (hit or miss). We need to find the probability of a specific number of successes (exactly 4 hits). This is a classic application of the Binomial Distribution. Step 2: Key Formula or Approach:
The binomial probability formula gives the probability of getting exactly \(k\) successes in \(n\) independent trials:
\[
P(X=k) = \binom{n}{k} p^k q^{n-k}
\]
where:
- \(n\) is the total number of trials.
- \(k\) is the number of successes.
- \(p\) is the probability of success in one trial.
- \(q = 1-p\) is the probability of failure in one trial. Step 3: Detailed Explanation: (i) Identify the parameters:
- Total number of trials (shots), \(n = 5\).
- Number of desired successes (hits), \(k = 4\).
- Probability of success (hitting the target), \(p = \frac{3}{4}\).
- Probability of failure (missing the target), \(q = 1 - p = 1 - \frac{3}{4} = \frac{1}{4}\). (ii) Apply the binomial formula:
\[
P(X=4) = \binom{5}{4} \left(\frac{3}{4}\right)^4 \left(\frac{1}{4}\right)^{5-4}
\]
\[
P(X=4) = \binom{5}{4} \left(\frac{3}{4}\right)^4 \left(\frac{1}{4}\right)^1
\]
(iii) Calculate the components and the final probability:
- \(\binom{5}{4} = \frac{5!}{4!(5-4)!} = \frac{5 \times 4!}{4! \times 1!} = 5\).
- \(\left(\frac{3}{4}\right)^4 = \frac{3^4}{4^4} = \frac{81}{256}\).
- \(\left(\frac{1}{4}\right)^1 = \frac{1}{4}\).
Now, multiply these parts together:
\[
P(X=4) = 5 \times \frac{81}{256} \times \frac{1}{4} = \frac{5 \times 81}{256 \times 4} = \frac{405}{1024}
\]
Step 4: Final Answer:
The probability of hitting the target exactly 4 times in 5 shots is \(\frac{405}{1024}\).