Question

Suppose n = 10 pistons are randomly selected. A manufacturer of metal pistons finds that on the average, 12% of his pistons are rejected because they are either oversize or undersize. Find the probability that at most two of the selected pistons are rejects.

Answer 1

n = 10

P(rejection) = 0.12

P(x <=2) = P(0) + P(1) + P(2)

$P(x) =\frac{ N!}{x! (N-x)!} \pi ^{x} (1-\pi)^{N-x}$

$P(0) =\frac{ 10 !}{0! (10-0)!} 0.12 ^{0} (1-0.12)^{10-0} = 0.2785$

$P(1) =\frac{ 10 !}{1! (10-1)!} 0.12 ^{1} (1-0.12)^{10-1} = 0.3798$

10! P(2) = 1:(10 27,0.12(1 – 0.12)10–2 = 0.2330

P(x <=2) = 0.8913

ANS : Probablity that at most two of the selected pistons are rejects = 0.8913