a production process produces 3% defective parts. a sample of ten parts from the production process is selected. what is the probability that the sample contains exactly three defective parts?



Answer :

The probability of a sample that contains exactly two defective parts is  13.1926.

As we know if P is the probability of achieving k results in n trials then the probability formula is P = [tex]({}^{n}_{k})[/tex][tex]p^{k}[/tex][tex]q^{n-k}[/tex]

In this formula n = number of trials

k = number of success

(n-k) = number of failures

p = probability of success in one trial

q = (1-p) = probability of failure in one trial

In this sum n = 10

k = 3

number failures (n-k) = (10 - 3) = 7

p = 3% which can be written as 0.03

q = 97% Which can be written as 0.97

Now putting these values in the formula

P = [tex]({}^{10}_{3})[/tex][tex]0.03^{3}[/tex][tex]0.97^{10-3}[/tex]

P = [tex]({}^{10}_{3})[/tex][tex]0.03^{3}[/tex][tex]0.97^{7}[/tex]

P = (10! ÷ 3!) × 0.000027 × 0.8079

P = 13.1926

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