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

Solution

Maths sir · 3 steps

Add the cases 0, 1 and 2, then pull (3/4)³ out of every term, and k is what is left.

1. Set up binomial. Each machine is out of service or not, independently, so the number out of service X is binomial with n = 5, p = 1/4 and q = 3/4. At most two means X = 0, 1 or 2.

   P(X ≤ 2) = q⁵ + C(5, 1) p q⁴ + C(5, 2) p² q³ = q³(q² + 5pq + 10p²)

2. Fill the bracket. q² = 9/16, 5pq = 5 × 1/4 × 3/4 = 15/16 and 10p² = 10/16. The bracket adds up to (9 + 15 + 10)/16 = 34/16 = 17/8.
3. So. P(X ≤ 2) = (3/4)³ × 17/8, which gives k = 17/8. That is option (c).

The idea to keep. For an at most or at least sum of binomial terms, factor out the lowest power of q shared by all the terms. The bracket that is left is usually the answer they want.

Answer from the NTA final answer key. Solution by PrepWiser.