Probability of a shooter of hitting the target is `(3)/(4)`. If he shoots 10 times, find the probability of hitting 8 targets successfully.

Answer

To solve the problem, we need to find the probability of a shooter hitting the target 8 times out of 10 shots, given that the probability of hitting the target is \( \frac{3}{4} \).

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Let \( n = 10 \) (total number of shots).
- Let \( p = \frac{3}{4} \) (probability of hitting the target).
- Let \( q = 1 - p = 1 - \frac{3}{4} = \frac{1}{4} \) (probability of missing the target).
- We want to find the probability of hitting the target \( r = 8 \) times.

2. **Use the Binomial Probability Formula**:
The probability of getting exactly \( r \) successes in \( n \) trials is given by the formula:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
where \( \binom{n}{r} \) is the binomial coefficient.

3. **Calculate the Binomial Coefficient**:
\[ \binom{10}{8} = \binom{10}{2} = \frac{10!}{8!(10-8)!} = \frac{10 \times 9}{2 \times 1} = 45 \]

4. **Substitute Values into the Formula**:
Now substituting \( n = 10 \), \( r = 8 \), \( p = \frac{3}{4} \), and \( q = \frac{1}{4} \) into the formula:
\[ P(X = 8) = \binom{10}{8} \left(\frac{3}{4}\right)^8 \left(\frac{1}{4}\right)^{10-8} \]
\[ P(X = 8) = 45 \left(\frac{3}{4}\right)^8 \left(\frac{1}{4}\right)^2 \]

5. **Calculate the Powers**:
\[ P(X = 8) = 45 \left(\frac{3^8}{4^8}\right) \left(\frac{1}{16}\right) \]
\[ P(X = 8) = 45 \cdot \frac{3^8}{4^{10}} \]

6. **Final Expression**:
Thus, the probability of hitting the target 8 times out of 10 is:
\[ P(X = 8) = \frac{45 \cdot 3^8}{4^{10}} \]