Question

a manager claims that 8% of part supplies purchased by his firm must be returned cause they are defective. if an order of 38 parts is received, what is the probability that 2 or 3 or 4 of these parts are defective?

Answer 1

Probability of defect (p) = 8% = 0.08

Probability of no defect (q) = 1 - p = 1 - 0.08 = 0.92

Sample size (N) = 38

Using binomial theorem,

P(X = x) = ^NC_x x p^x x q^{(N - x)} = ³⁸C_x x (0.08)^x x (0.92)^{(38 - x)}, where

^NC_x = (N!) / [(x!) x {(N - x)!}]

Therefore,

P(X = 2) + P(X = 3) + P(X = 4) = ³⁸C₂ x (0.08)² x (0.92)^{(38 - 2)} + ³⁸C₃ x (0.08)³ x (0.92)^{(38 - 3)} + ³⁸C₄ x (0.08)⁴ x (0.92)^{(38 - 4)}

= 703x 0.0064 x (0.92)⁽³⁶⁾ + 8436 x 0.000512 x (0.92)⁽³⁵⁾ + 73815x 0.00004096 x (0.92)⁽³⁴⁾

= 0.2236 + 0.2333 + 0.1775

= 0.6344