Answer:
P(X ≥ 2) = 0.5967
Step-by-step explanation:
Using binomial probability distribution formula;
P(X = k) = C(n, k) × p^(k) × (1 - p)^(n - k)
two percent of a product are defective. Thus; p = 0.02
a lot of 100 items are ordered. Thus, n = 100
Probability that there are no defective items is;
P(X = 0) = C(100, 0) × 0.02^(0) × (1 - 0.02)^(100 - 0)
P(X = 0) = 0.98^(100) = 0.1326
the probability that there are at least two defective items will be;
P(X ≥ 2) = 1 - P(X < 2)
Now,
P(X < 2) = P(X = 0) + P(X = 1)
P(X = 1) = C(100, 1) × 0.02^(1) × (1 - 0.02)^(100 - 1)
P(X = 1) = 100 × 0.02 × 0.98^(99) = 0.2707
Thus;
P(X < 2) = 0.1326 + 0.2707
P(X < 2) = 0.4033
P(X ≥ 2) = 1 - 0.4033
P(X ≥ 2) = 0.5967
