Question

A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historically, the failure rate for LED light bulbs that the company manufactures is 11%. Suppose a random sample of 10 LED light bulbs is selected. What is the probability that a) None of the LED light bulbs are defective? b) Exactly one of the LED light bulbs is defective? c) Two or fewer of the LED light bulbs are defective? d) Three or more of the LED light bulbs are not defective?

Answer 1

a)
Here, n = 10, p = 0.11, (1 - p) = 0.89 and x = 0
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 0)
P(X = 0) = 10C0 * 0.11^0 * 0.89^10
P(X = 0) = 0.3118

b)
We need to calculate P(X = 1)
P(X = 1) = 10C1 * 0.11^1 * 0.89^9
P(X = 1) = 0.3854

c)
P(X <= 2) = (10C0 * 0.11^0 * 0.89^10) + (10C1 * 0.11^1 * 0.89^9) + (10C2 * 0.11^2 * 0.89^8)
P(X <= 2) = 0.3118 + 0.3854 + 0.2143
P(X <= 2) = 0.9115

d)
Here, n = 10, p = 0.89, (1 - p) = 0.11 and x = 2
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 2).
P(X <= 2) = (10C0 * 0.89^0 * 0.11^10) + (10C1 * 0.89^1 * 0.11^9) + (10C2 * 0.89^2 * 0.11^8)
P(X <= 2) = 0 + 0 + 0
P(X <= 2) = 0

P(X >= 3) = 1 - 0 = 1