Single Choice

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

A$$4$$
B$$\dfrac{17}{2}$$
C$$\dfrac{17}{8}$$
Correct Answer
D$$\dfrac{17}{4}$$

Solution

Given $$n=5,p=\dfrac 14,q=1-p=\dfrac 34$$ Required probability = when no machine has fault + when only one machine has fault + when only two machine has fault $$= \, ^5C_0 \left(\dfrac{3}{4} \right)^5 + \,^5C_1 \left(\dfrac{1}{4} \right) \left(\dfrac{3}{4} \right)^4 + \, ^5C_2 \left(\dfrac{1}{4} \right)^2 \left(\dfrac{3}{4} \right)^3$$ $$= \dfrac{243}{1024} + \dfrac{405}{1024} + \dfrac{270}{1024}$$ $$= \dfrac{459}{512}$$ $$= \left(\dfrac{3}{4} \right)^3 \times k = \left(\dfrac{3}{4} \right)^3 \times \dfrac{17}{8}$$ by comparing we get $$k = \dfrac{17}{8}$$

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