1. Key Concepts and Formulas

• Binomial Probability Distribution: This distribution is used to model the number of "successes" in a fixed number of independent trials, where each trial has only two possible outcomes (success or failure) and the probability of success is constant for each trial.
• Probability Mass Function (PMF): For a binomial distribution, the probability of getting exactly $r$ successes in $n$ trials is given by: $P(X=r) = \binom{n}{r} p^r q^{n-r}$ where:
  • $X$ is the random variable representing the number of successes.
  • $n$ is the total number of trials.
  • $r$ is the number of successes of interest.
  • $p$ is the probability of success in a single trial.
  • $q$ is the probability of failure in a single trial, where $q = 1 - p$.
  • $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ is the binomial coefficient.

2. Step-by-Step Solution

Step 1: Identify Parameters and Define the Random Variable We begin by extracting the given information from the problem to define the parameters for our binomial distribution.

• Total number of machines ($n$): There are five machines, so $n=5$.
• Probability of a machine being out of service ($p$): This is our "probability of success," given as $p = \frac{1}{4}$.
• Probability of a machine being operational ($q$): This is our "probability of failure," calculated as $q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}$.

Let $X$ be the random variable representing the number of machines that are out of service on a given day. We are asked to find the probability that "at most two machines will be out of service," which translates to $P(X \le 2)$. This means we need to sum the probabilities for $X=0$, $X=1$, and $X=2$.

Step 2: Calculate Probabilities for Each Case ($X=0, X=1, X=2$) We use the binomial probability formula $P(X=r) = \binom{n}{r} p^r q^{n-r}$ for each required value of $r$.

• Case 1: Exactly 0 machines out of service ($r=0$) This means all five machines are operational. $P(X=0) = \binom{5}{0} \left(\frac{1}{4}\right)^0 \left(\frac{3}{4}\right)^{5-0}$ Since $\binom{5}{0} = 1$ and any non-zero number raised to the power of 0 is 1: $P(X=0) = 1 \times 1 \times \left(\frac{3}{4}\right)^5 = \left(\frac{3}{4}\right)^5$

• Case 2: Exactly 1 machine out of service ($r=1$) This means one machine is out of service and the other four are operational. $P(X=1) = \binom{5}{1} \left(\frac{1}{4}\right)^1 \left(\frac{3}{4}\right)^{5-1}$ Since $\binom{5}{1} = 5$: $P(X=1) = 5 \times \left(\frac{1}{4}\right)^1 \times \left(\frac{3}{4}\right)^4 = \frac{5}{4} \left(\frac{3}{4}\right)^4$

• Case 3: Exactly 2 machines out of service ($r=2$) This means two machines are out of service and the other three are operational. $P(X=2) = \binom{5}{2} \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^{5-2}$ First, calculate the binomial coefficient: $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10$. $P(X=2) = 10 \times \left(\frac{1}{4}\right)^2 \times \left(\frac{3}{4}\right)^3 = 10 \times \left(\frac{1}{16}\right) \times \left(\frac{3}{4}\right)^3 = \frac{10}{16} \left(\frac{3}{4}\right)^3 = \frac{5}{8} \left(\frac{3}{4}\right)^3$

Step 3: Calculate the Total Probability $P(X \le 2)$ The probability that at most two machines will be out of service is the sum of the probabilities calculated in Step 2: $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$ Substitute the expressions: $P(X \le 2) = \left(\frac{3}{4}\right)^5 + \frac{5}{4} \left(\frac{3}{4}\right)^4 + \frac{5}{8} \left(\frac{3}{4}\right)^3$ To simplify and match the given form ${\left( {{3 \over 4}} \right)^3}k$, we factor out the common term $\left(\frac{3}{4}\right)^3$: $P(X \le 2) = \left(\frac{3}{4}\right)^3 \left[ \left(\frac{3}{4}\right)^2 + \frac{5}{4} \left(\frac{3}{4}\right)^1 + \frac{5}{8} \right]$ Now, simplify the terms inside the square brackets: $P(X \le 2) = \left(\frac{3}{4}\right)^3 \left[ \frac{9}{16} + \frac{15}{16} + \frac{5}{8} \right]$ To add these fractions, we find a common denominator, which is 16: $P(X \le 2) = \left(\frac{3}{4}\right)^3 \left[ \frac{9}{16} + \frac{15}{16} + \frac{5 \times 2}{8 \times 2} \right]$ $P(X \le 2) = \left(\frac{3}{4}\right)^3 \left[ \frac{9}{16} + \frac{15}{16} + \frac{10}{16} \right]$ Add the numerators: $P(X \le 2) = \left(\frac{3}{4}\right)^3 \left[ \frac{9 + 15 + 10}{16} \right]$ $P(X \le 2) = \left(\frac{3}{4}\right)^3 \left[ \frac{34}{16} \right]$ Simplify the fraction $\frac{34}{16}$ by dividing both numerator and denominator by 2: $P(X \le 2) = \left(\frac{3}{4}\right)^3 \left[ \frac{17}{8} \right]$

Step 4: Solve for k The problem states that the probability that at most two machines will be out of service is equal to ${\left( {{3 \over 4}} \right)^3}k$. We have calculated this probability to be $\left(\frac{3}{4}\right)^3 \left[ \frac{17}{8} \right]$. Equating the two expressions: ${\left( {{3 \over 4}} \right)^3}k = \left(\frac{3}{4}\right)^3 \left[ \frac{17}{8} \right]$ By comparing both sides, we can clearly see that: $k = \frac{17}{8}$

3. Common Mistakes & Tips

• Misinterpreting "At Most": "At most two" means $X \le 2$, including $X=0, X=1, X=2$. A common error is to only calculate for $X=2$.
• Binomial Coefficient Errors: Ensure accurate calculation of $\binom{n}{r}$. Forgetting this factor or calculating it incorrectly is a frequent mistake.
• Algebraic Simplification: Be meticulous with fraction arithmetic and factoring out common terms. Factoring $\left(\frac{3}{4}\right)^3$ early significantly simplifies the calculation.

4. Summary

This problem is a direct application of the Binomial Probability Distribution. We identified the number of trials ($n=5$), the probability of success ($p=1/4$), and the probability of failure ($q=3/4$). We then calculated the probabilities for 0, 1, and 2 machines being out of service using the binomial PMF. Summing these probabilities gave us $P(X \le 2) = \left(\frac{3}{4}\right)^3 \left[ \frac{17}{8} \right]$. By equating this to the given expression ${\left( {{3 \over 4}} \right)^3}k$, we found the value of $k$.

The final answer is \boxed{\text{{{{17} \over 8}}}}, which corresponds to option (C).