Question

The probability that a part manufactured by a company will be defective is 0.05. If 15 such parts are selected randomly and inspected, then the probability that at least two parts will be defective is ______ (round off to two decimal places).

Answer 1

Concept:

Binomial probability distribution: \(P\left( x \right) = {}_{}^n{C_x}{p^x}{q^{n - x}} = \frac{{n!}}{{\left( {n - x} \right)! \times x!}}{p^x}{q^{n - x}}\)

Probability of defective (P_d) = p = 0.05

q = 1 – p = 0.95

15 parts are selected at random i.e. n = 15

To find probability that at least two defected

So, this is equivalent to find more than 0 or 1 defective

⇒ Probability of 0 defective = P (0) = ¹⁵C₀p⁰q¹⁵ = 1 × 0.95¹⁵ = 0.4633.

Probability of 1 defective = P (1) = ¹⁵C₁p¹q¹⁴ = 15 × 0.05¹ × 0.95¹⁴ = 0.3657

⇒ Probability of 0 or 1 defective = 0.4633 + 0.3657 = 0.829

⇒ P (defective ≥ 2) = 1 – P (defective ≤ 1)

⇒ P = 1 – 0.829 = 0.171

P = 0.17 (Rounded to two decimals).