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Question
If 18% of the bolts produced by a machine are defective, determine the probability that out of the 4 bolts chosen at random exactly one will be defective
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Solution
n = 4
Probability of defective bolts p = `18/100`
9 = 1 – p = 1.018 = 0.82
The binoial distribution p(x) = 4Cx(0.18)x(0.82)4-x
= 4C1(0.18)1(0.82)4-1
= 4 × 0.18 × (0.82)3
= 0.72 × 0.551368
= 0.39698496
P(X = 1)= 0.3969 approximately
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