Question

Assume that 20 parts are checked each hour and that X denotes the number of parts in the sample of 20 that require rework. Parts are assumed to be independent with respect to rework.

If the percentage of parts that require rework remains at 1%, what is the probability that hour 10 is the 2nd sample at which X exceeds 1?

Answer 1

X ~ Binomial(n = 20, p = 0.01)

P(X > 1) = 1 - P(X 1) = 1 - [P(X = 0) + P(X = 1)]

= 1 - [²⁰C₀ * 0.01⁰ * (1 - 0.01)^20-0 + ²⁰C₁ * 0.01¹ * (1 - 0.01)^20-1]

= 1 - (0.99²⁰ + 20 * 0.01 * 0.99¹⁹ )

= 0.01685934

Let Y be the number of hours in which X exceeds 1. Then Y ~ NegativeBinomial(r = 2, p = 0.01685934)

The PMF of Y is,

P(Y = y) = ^y-1C_r-1 * p^r * (1 - p)^y-r = ^y-1C_2-1 * 0.01685934² * (1 - 0.01685934)^y-2

P(Y = 10) = ⁹C₁ * 0.01685934² * (1 - 0.01685934)⁸

= 9 * 0.01685934² * 0.9831407⁸

= 0.002232796