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An inspector samples four PC’s from a steady stream of computers that is known to be 12% nonconforming. What is the probability of selecting two nonconforming units in the sample? Write your answer to 3 decimal places in the format: x.xxx

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Answer: 0.067

Explanation:

Binomial distribution formula :-


P(X)=^nC_xp^x(1-p)^(n-x), here P(x) is the probability of getting success in x trials , n is total number of trials and p is the probability of getting success in each trial.

Given : n = 4 , p = 12 % = 0.12 and x =2

Then ,
P(2)=^4C_2(0.12)^2(1-0.12)^(4-2)


\Rightarrow\ P(2)=(4!)/(2!2!)(0.12)^2(0.88)^2=0.06690816\approx0.067

∴ The probability of selecting two nonconforming units in the sample = 0.067

User Serge Kvashnin
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