The probability that a man can hit a target is `3//4`. He tries 5 times. The probability that he will hit the target atleast three times, is

Options

A. `291/364`
B. `371/464`
C. `471/502`
D. `459/512`

Answer

To solve the problem, we need to find the probability that a man hits a target at least 3 times when he tries 5 times, given that the probability of hitting the target in a single attempt is \( \frac{3}{4} \).

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Probability of hitting the target, \( p = \frac{3}{4} \)
- Probability of missing the target, \( q = 1 - p = \frac{1}{4} \)
- Number of trials, \( n = 5 \)

2. **Define the Event**:
- We want to find the probability of hitting the target at least 3 times. This means we need to calculate the probabilities for hitting the target exactly 3 times, exactly 4 times, and exactly 5 times.

3. **Use the Binomial Probability Formula**:
The binomial probability formula is given by:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
where \( \binom{n}{k} \) is the binomial coefficient, \( p \) is the probability of success, \( q \) is the probability of failure, \( n \) is the number of trials, and \( k \) is the number of successes.

4. **Calculate the Probabilities**:
- **For \( k = 3 \)**:
\[ P(X = 3) = \binom{5}{3} \left(\frac{3}{4}\right)^3 \left(\frac{1}{4}\right)^{5-3} \]
\[ = 10 \cdot \left(\frac{27}{64}\right) \cdot \left(\frac{1}{16}\right) = 10 \cdot \frac{27}{1024} = \frac{270}{1024} \]

- **For \( k = 4 \)**:
\[ P(X = 4) = \binom{5}{4} \left(\frac{3}{4}\right)^4 \left(\frac{1}{4}\right)^{5-4} \]
\[ = 5 \cdot \left(\frac{81}{256}\right) \cdot \left(\frac{1}{4}\right) = 5 \cdot \frac{81}{1024} = \frac{405}{1024} \]

- **For \( k = 5 \)**:
\[ P(X = 5) = \binom{5}{5} \left(\frac{3}{4}\right)^5 \left(\frac{1}{4}\right)^{5-5} \]
\[ = 1 \cdot \left(\frac{243}{1024}\right) \cdot 1 = \frac{243}{1024} \]

5. **Sum the Probabilities**:
Now, we add the probabilities for hitting the target at least 3 times:
\[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) \]
\[ = \frac{270}{1024} + \frac{405}{1024} + \frac{243}{1024} = \frac{270 + 405 + 243}{1024} = \frac{918}{1024} \]

6. **Simplify the Result**:
To simplify \( \frac{918}{1024} \):
- The greatest common divisor (GCD) of 918 and 1024 is 2.
\[ \frac{918 \div 2}{1024 \div 2} = \frac{459}{512} \]

### Final Answer:
The probability that he will hit the target at least 3 times is \( \frac{459}{512} \).