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 $\frac{17}{8}$

P(machine being faulty) $=p=\frac{1}{4}$
$\therefore$P(machine not being faulty)$=q=1-p=\frac{3}{4}$
P(at most two machines being faulty)
$=$ P(zero machine being faulty) $+$
P(one machine being faulty) $+$
P(two machines being faulty)

$={}^5{C}_0{p}^0{q}^5+{}^5{C}_1{p}^1{q}^4+{}^5{C}_2{p}^2{q}^3$
$=q^5+5p{q}^4+10p^2{q}^3$
$={\left(\frac{3}{4}\right)}^5+5\times \frac{1}{4}{\left(\frac{3}{4}\right)}^4+10\times {\left(\frac{1}{4}\right)}^2{\left(\frac{3}{4}\right)}^3$
$={\left(\frac{3}{4}\right)}^3[\frac{9}{16}+\frac{15}{16}+\frac{10}{16}]$
$={\left(\frac{3}{4}\right)}^3\times \frac{34}{16}={\left(\frac{3}{4}\right)}^3\times \frac{17}{8}$

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