two percent of a product are defective. if a lot of 100 items are ordered what's the probability that there are no defective item? what's the probability that there are at least two defective items?

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

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