10% of electric light bulbs are defective at manufacturer. A random sample of 8 light bulbs are chosen and tested. Find the probability that at least 3 are defective.

0.0381

Step-by-step explanation:

This is a binomial probability distribution problem. Thus;

p = 10% = 0.1

We are told that A random sample of 8 light bulbs are chosen and tested. Thus, n = 8

We want to find the probability that at least 3 are defective. This will be;

P(X ≥ 3) = P(3) + P(4) + P(5) + P(6) + P(7) + P(8)

P(X = x) = C(n, x) × p^(x) × (1 - p)^(n - x)

Thus;

P(3) = C(8, 3) × 0.1³ × (1 - 0.1)^(8 - 3)

P(3) = C(8, 3) × 0.1³ × 0.9^(5)

P(3) = 0.03306744

Repeating this same procedure, we have;

P(4) = 0.0045927

P(5) = 0.00040824

P(6) = 0.00002268

P(7) = 0.00000072

P(8) = 0.00000001

Thus;

P(X ≥ 3) = 0.03306744 + 0.0045927 + 0.00040824 + 0.00002268 + 0.00000072 + 0.00000001 = 0.03809179

Approximately 0.0381

the answer to your question is b.