GATE Mechanical Engineering (ME) 2019 Shift 2 Solved Paper

Question : 47 of 65

Marks: +1, -0

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).

Your Answer:

Validate

Solution:

Total no. of parts = 15
P[defective(p)] = 0.05
P[non-defective(q)] = 1 – 0.05 = 0.95
P[at least 2 defective] = 1 – [P(no defective)+P(1 defective)]
Using the binomial distribution

{P}[

at least 2 defective]

=1-[{ }^{15} {C}_{0} {p}^{0} {q}^{15}+{ }^{15} {C}_{1} {p}^{1} {q}^{14}]=1-[0.05^{∘} 0.95^{15}+15 × 0.05 × 0.95^{14}]=1-

[0.4639+0.3657]=1-0.8296=0.1704

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