Answer:
a) 0.3585
b) 0.6415
c) 0.07548
Step-by-step explanation:
This is a binomial distribution problem
Binomial distribution function is represented by
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
n = total number of sample spaces = 20
x = Number of successes required
p = probability of window frame with minor defect = 0.05
q = probability of window frame with no defect = 1 - 0.05 = 0.95
a) Probability that none of the frames picked will have defects
x = 0
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
P(X = 0) = ²⁰C₀ 0.05⁰ 0.95²⁰⁻⁰ = 0.3585
b) Probability that At least one will need adjustment = P(X ≥ 1) = 1 - P(X < 1) = 1 - P(X = 0) = 1 - 0.3585 = 0.6415
c) Probability that More than two will need adjustment = P(X > 2)
P(X > 2) = 1 - P(X ≤ 2) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
So, we evaluate each of that using the binomial distribution formula, slotting in the appropriate variables, and sum it all up.
P(X > 2) = 1 - 0.9245 = 0.07548
