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Question
A manufacturer of metal pistons finds that on the average, 12% of his pistons are rejected because they are either oversize or undersize. What is the probability that a batch of 10 pistons will contain no more than 2 rejects?
Solution
In a binomial distribution
n = 10
p = `12/100 = 3/25`
q = – p = `1 - 3/25` q = = `22/25`
PX = x) = ncxpxqn-x
P(no more than 2 rejects)
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= `""^10"C"_0 (3/25)^0 (22/25)^(10 - 0) + ""^10"C"_1 (3/25)^1 (22/25)^(10 - 1) + ""^10"C"_2 (3/25)^2 + (22/25)^(10 - 2)`
= `(1)(1) (22/25)^10 + 10 ((3 xx (22)^9)/((25)^10)) + (10 xx 9)/(1 xx 2) (3/25)^2 (22/25)^8`
= `((22)^10 + (30 xx (22)^9) + 405(22)^8)/(25)^10`
= `((22)^8 [(22)^2 + (30 xx 22) + 405])/(25)^10`
= `(5.486 xx 10^10 xx 1549)/(9.5288 xx 10^13`
= `(5.486 xx 1549)/(9.528 xx 10^3)`
= `(5.486 xx 1549)/9528`
= 0.89187
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