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 \( {{1 \over 4}} \) . If the probability that at most two machines will be out of service on the same day is \( {\left( {{3 \over 4}} \right)^3}k \), then k is equal to :

• A. \( {{{17} \over 4}} \)
• B. \( {{{17} \over 2}} \)
• C. \( {{{17} \over 8}} \)
• D. 4

Hint:

Use the binomial distribution with n = 5 and p = 1/4; add the probabilities for 0, 1 and 2 machines out of service.

Answer 1

The Correct Option is C

Step 1: \(P(X\le2)=\binom50\left(\tfrac34\right)^5+\binom51\tfrac14\left(\tfrac34\right)^4+\binom52\left(\tfrac14\right)^2\left(\tfrac34\right)^3\).
Step 2: Factor \(\left(\tfrac34\right)^3\): \(\left(\tfrac34\right)^3\left[\tfrac9{16}+5\cdot\tfrac14\cdot\tfrac34+\tfrac{10}{16}\right]=\left(\tfrac34\right)^3\left[\tfrac9{16}+\tfrac{15}{16}+\tfrac{10}{16}\right]=\left(\tfrac34\right)^3\cdot\tfrac{34}{16}\).
Step 3: Hence \(k=\tfrac{34}{16}=\tfrac{17}8\).
Answer: (C) \(\tfrac{17}{8}\)