In a workshop, there are five machines and the probability of any one of them to be out of service on a day is ¼. If the probability that at most two machines will be out of service on the same day is 343k, then k is equal to

Solution:

Given, probability of any machine to be out of service $= \frac{1}{4}$
So, probability of no machine is out of service $= 1 - \frac{1}{4} = \frac{3}{4}$

Required probability = when no machine is out of service + when only one machine is out of service + when only two machines are out of service.

$=^{5} C_{0} {(\frac{3}{4})}^{5} +^{5} C_{1} (\frac{1}{4}) {(\frac{3}{4})}^{4} +^{5} C_{2} {(\frac{1}{4})}^{2} {(\frac{3}{4})}^{3}$

$= {(\frac{3}{4})}^{3} [\frac{9}{16} + 5 (\frac{3}{16}) + 10 (\frac{1}{16})]$

$= {(\frac{3}{4})}^{3} (\frac{9 + 15 + 10}{16}) = {(\frac{3}{4})}^{3} (\frac{34}{16}) = {(\frac{3}{4})}^{3} (\frac{17}{8}) = {(\frac{3}{4})}^{3} k$

$∴ k = \frac{17}{8}$

In a workshop, there are five machines and the probability of any one of them to be out of service on a day is ¼. If the probability that at most two machines will be out of service on the same day is ${(\frac{3}{4})}^{3} k$ , then k is equal to

Solution:

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