In a workshop, there are five machines and the probability of any one of them to be out of service on a day is $$\frac{1}{4}$$. If the probability that at most two machines will be out of service on the same day is $$\left(\frac{3}{4}\right)^3 k$$, then $$k$$ is equal to

We are told that each of the five machines fails independently and that for any single machine the probability of being out of service on a particular day is $$\frac14$$. Therefore, for one machine the probability of being in service is $$1-\frac14=\frac34$$.

If we let the random variable $$X$$ denote the number of machines that are out of service on a given day, then $$X$$ follows a binomial distribution with parameters $$n=5$$ and $$p=\frac14$$. The general binomial probability formula is

$$P(X=r)=\binom{n}{r}\,p^{\,r}\,(1-p)^{\,n-r}.$$

Here the phrase “at most two machines will be out of service’’ means the event $$X\le 2$$, i.e. $$X=0,1,2$$. Thus we need

$$P(X\le 2)=P(X=0)+P(X=1)+P(X=2).$$

We now evaluate each of these three terms one by one.

For $$r=0$$ machines out of service we have

$$P(X=0)=\binom{5}{0}\Bigl(\tfrac14\Bigr)^{0}\Bigl(\tfrac34\Bigr)^{5} =1\cdot1\cdot\Bigl(\tfrac34\Bigr)^{5} =\Bigl(\tfrac34\Bigr)^{5}.$$

For $$r=1$$ machine out of service we have

$$P(X=1)=\binom{5}{1}\Bigl(\tfrac14\Bigr)^{1}\Bigl(\tfrac34\Bigr)^{4} =5\cdot\Bigl(\tfrac14\Bigr)\cdot\Bigl(\tfrac34\Bigr)^{4}.$$

For $$r=2$$ machines out of service we have

$$P(X=2)=\binom{5}{2}\Bigl(\tfrac14\Bigr)^{2}\Bigl(\tfrac34\Bigr)^{3} =10\cdot\Bigl(\tfrac14\Bigr)^{2}\cdot\Bigl(\tfrac34\Bigr)^{3}.$$

Adding these three results gives the total probability:

$$ \begin{aligned} P(X\le 2) &=\Bigl(\tfrac34\Bigr)^{5} +5\Bigl(\tfrac14\Bigr)\Bigl(\tfrac34\Bigr)^{4} +10\Bigl(\tfrac14\Bigr)^{2}\Bigl(\tfrac34\Bigr)^{3}.\\ \end{aligned} $$

Because the final answer must be expressed in the form $$\bigl(\tfrac34\bigr)^{3}k$$, we factor $$\bigl(\tfrac34\bigr)^{3}$$ from every term:

$$ \begin{aligned} P(X\le 2) &=\Bigl(\tfrac34\Bigr)^{3} \left[ \Bigl(\tfrac34\Bigr)^{2} +5\Bigl(\tfrac14\Bigr)\Bigl(\tfrac34\Bigr) +10\Bigl(\tfrac14\Bigr)^{2} \right]. \end{aligned} $$

We now simplify each bracketed expression separately.

First term inside brackets:

$$\Bigl(\tfrac34\Bigr)^{2}=\frac{9}{16}.$$

Second term inside brackets:

$$5\Bigl(\tfrac14\Bigr)\Bigl(\tfrac34\Bigr) =5\cdot\frac14\cdot\frac34 =5\cdot\frac{3}{16} =\frac{15}{16}.$$

Third term inside brackets:

$$10\Bigl(\tfrac14\Bigr)^{2} =10\cdot\frac{1}{16} =\frac{10}{16} =\frac{5}{8}.$$

Now we add these three fractions, being careful to use a common denominator of 16:

$$ \begin{aligned} \frac{9}{16} + \frac{15}{16} + \frac{5}{8} &=\frac{9}{16} + \frac{15}{16} + \frac{10}{16}\\ &=\frac{9+15+10}{16}\\ &=\frac{34}{16}\\ &=\frac{17}{8}. \end{aligned} $$

Therefore

$$P(X\le 2)=\Bigl(\tfrac34\Bigr)^{3}\cdot\frac{17}{8}.$$

Comparing with the required form $$\Bigl(\frac34\Bigr)^{3}k$$ we see directly that

$$k=\frac{17}{8}.$$

Hence, the correct answer is Option A.