There are 5 machines. Probability of a machine being faulted is `1/4` . Probability of atmost two machines is faulted, is `(3/4)^3 k`, then value of k is

Options

A. 1. `(17)/(2)`
B. 2. `(17)/(8)`
C. 3. 4
D. 4.`(17)/(4)`

Answer

To solve the problem, we need to calculate the probability of at most two machines being faulted out of five machines, where the probability of a machine being faulted is \( \frac{1}{4} \).

### Step-by-Step Solution:

1. **Identify Probabilities**:
- Let \( P(F) = \frac{1}{4} \) (probability of a machine being faulted).
- Let \( P(NF) = 1 - P(F) = 1 - \frac{1}{4} = \frac{3}{4} \) (probability of a machine not being faulted).

2. **Define the Problem**:
- We need to find the probability of at most 2 machines being faulted. This includes the cases where:
- No machines are faulted (0 faulted).
- Exactly 1 machine is faulted (1 faulted).
- Exactly 2 machines are faulted (2 faulted).

3. **Use the Binomial Probability Formula**:
- The probability of exactly \( r \) successes (faulted machines) in \( n \) trials (total machines) is given by:
\[ P(X = r) = \binom{n}{r} (P(F))^r (P(NF))^{n-r} \]
- Here, \( n = 5 \) (total machines), \( P(F) = \frac{1}{4} \), and \( P(NF) = \frac{3}{4} \).

4. **Calculate Each Case**:
- **Case 0 (No machines faulted)**:
\[ P(X = 0) = \binom{5}{0} \left(\frac{1}{4}\right)^0 \left(\frac{3}{4}\right)^5 = 1 \cdot 1 \cdot \left(\frac{3}{4}\right)^5 = \left(\frac{3}{4}\right)^5 \]
- **Case 1 (One machine faulted)**:
\[ P(X = 1) = \binom{5}{1} \left(\frac{1}{4}\right)^1 \left(\frac{3}{4}\right)^4 = 5 \cdot \left(\frac{1}{4}\right) \cdot \left(\frac{3}{4}\right)^4 = 5 \cdot \frac{1}{4} \cdot \frac{81}{256} = \frac{405}{1024} \]
- **Case 2 (Two machines faulted)**:
\[ P(X = 2) = \binom{5}{2} \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^3 = 10 \cdot \left(\frac{1}{16}\right) \cdot \left(\frac{27}{64}\right) = 10 \cdot \frac{27}{1024} = \frac{270}{1024} \]

5. **Sum the Probabilities**:
- Now, we add the probabilities from the three cases:
\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = \left(\frac{3}{4}\right)^5 + \frac{405}{1024} + \frac{270}{1024} \]
- Calculate \( \left(\frac{3}{4}\right)^5 \):
\[ \left(\frac{3}{4}\right)^5 = \frac{243}{1024} \]
- Now, combine:
\[ P(X \leq 2) = \frac{243}{1024} + \frac{405}{1024} + \frac{270}{1024} = \frac{243 + 405 + 270}{1024} = \frac{918}{1024} \]

6. **Express in the Required Form**:
- We need to express this in the form \( \left(\frac{3}{4}\right)^3 k \):
\[ \frac{918}{1024} = \left(\frac{3}{4}\right)^3 k \]
- Calculate \( \left(\frac{3}{4}\right)^3 = \frac{27}{64} \).
- Thus, we have:
\[ \frac{918}{1024} = \frac{27}{64} k \]
- To find \( k \):
\[ k = \frac{918}{1024} \cdot \frac{64}{27} \]
- Simplifying:
\[ k = \frac{918 \cdot 64}{1024 \cdot 27} = \frac{918 \cdot 64}{1024 \cdot 27} = \frac{918}{1024} \cdot \frac{64}{27} = \frac{17}{8} \]

### Final Answer:
The value of \( k \) is \( \frac{17}{8} \).