Question

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

• A. $\frac{17}{2}$
• B. $4$
• C. $\frac{17}{4}$
• D. $\frac{17}{8}$

Answer 1

The correct option is D

$\frac{17}{8}$

Explanation for Correct Answer:

Use the binomial distribution method to find the probability:

Given that five machines are in the workshop.

Let the probability of the machine being faulty be given by,

$P\left(\text{machine being faulty}\right)=P=\frac{1}{4}$(given)
Therefore, the probability of the machine being non-faulty is given by:

$\begin{array}{l}q=1-\frac{1}{4}\\ =\frac{3}{4}\end{array}$.

Now,

$\begin{array}{l}P\left(\text{atmost 2 machines being faulty}\right)=P\left(\text{zero machine being faulty}\right)+P\left(\text{one machine being faulty}\right)+P\left(\text{two machine being faulty}\right)\\ ={}^5\mathrm{ℂ}_0{p}^0{q}^5+{}^5\mathrm{ℂ}_1{p}^1{q}^4+{}^5\mathrm{ℂ}_2{p}^2{q}^3\\ =\frac{5!}{(5-0)!\left(0\right)!}{q}^5+\frac{5!}{(5-1)!\left(1\right)!}{p}^1{q}^4+\frac{5!}{(5-2)!\left(3\right)!}{p}^2{q}^3\\ =1\times q^5+5\times p{q}^4+10\left(p^2{q}^3\right)\\ ={\left(\frac{3}{4}\right)}^5+5\times \left(\frac{1}{4}\right)\times {\left(\frac{3}{4}\right)}^4+10\times {\left(\frac{1}{4}\right)}^2\times {\left(\frac{3}{4}\right)}^3\\ ={\left(\frac{3}{4}\right)}^3\left[{\left(\frac{3}{4}\right)}^2+\frac{5\times 1\times 3}{4\times 4}+\frac{10}{16}\right]\\ ={\left(\frac{3}{4}\right)}^3\left[\frac{9}{16}+\frac{15}{16}+\frac{10}{16}\right]\\ ={\left(\frac{3}{4}\right)}^3\left(\frac{17}{8}\right)\end{array}$

Equate the obtained value with the given value

$\begin{array}{rcl}{\left(\frac{3}{4}\right)}^3\left(\frac{17}{8}\right)& =& {\left(\frac{3}{4}\right)}^{3}k\\ & \Rightarrow & k=\frac{17}{8}\end{array}$

Hence, option (D) is correct.