Question

A container holds 50 electronic components, of which 10 are defective.

Answer 1

1.) q = P(defective) = 10/50 = 0.2
p = P(not defective) = 1 - P(defective) = 1 - 0.2 = 0.8
P(x) = nCr p^x q^(n-x)
P(x ≥ 4) = P(4) + P(5) + P(6) = 6C4 * (0.8)^4 * (0.2)^2 + 6C5 * (0.8)^5 * 0.2 + (0.8)^6 = 15 * 0.4096 * 0.04 + 6 * 0.32768 * 0.2 + 0.262144 = 0.24576 + 0.393216 + 0.262144 = 0.90112
Option D is the correct answer.

2.) p = P(defective) = 10/50 = 0.2
q = P(not defective) = 1 - P(defective) = 1 - 0.2 = 0.8
P(x) = nCr p^x q^(n-x)
P(x = 3) = 8C3 * (0.2)^3 * (0.8)^5 = 8C3 * 0.008 * 0.32768 = 56 * 0.008 * 0.32768 = 0.1468
Option A is the correct answer.