JEE PYQ: Probability Question 26
Question 26 - 2020 (07 Jan Shift 2)
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:
(1) $\frac{17}{8}$
(2) $\frac{17}{4}$
(3) $\frac{17}{2}$
(4) $4$
Show Answer
Answer: (1)
Solution
$P(\text{at most 2}) = {}^5C_0\left(\frac{3}{4}\right)^5 + {}^5C_1\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^4 + {}^5C_2\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^3 = \left(\frac{3}{4}\right)^3\left[\frac{243+405+270}{1024}\right] = \left(\frac{3}{4}\right)^3 \times \frac{918}{1024}$. Simplifying: $= \left(\frac{3}{4}\right)^3 \times \frac{459}{512} = \left(\frac{3}{4}\right)^3 \times \frac{17}{8} \cdot …$. Actually: $\frac{243}{1024} + \frac{405}{1024} + \frac{270}{1024} = \frac{918}{1024}$. $\frac{918}{1024} = \left(\frac{3}{4}\right)^3 k \Rightarrow k = \frac{918}{1024 \times 27/64} = \frac{918 \times 64}{1024 \times 27} = \frac{918}{432} = \frac{17}{8}$.
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