Question

 A supplier delivers 25 parts in a factory. The supplier knows that 3 of the delivered items are defective. As always, the factory check the quality. In the end, 2 of the 25 components are randomly selected and thoroughly checked. The rule is that the goods will be refused if both examined goods don't meet the requirements. Calculate the probability that the supplier will be accepted?

Answer 1

Do i have to work with combination and Laplace?

Answer 2

for a part ... p(defective) = 3/25 = .12 ... p(not defective) = 1 - .12 = .88

this is a binary probability ... defective (d) or not defective (n)

(d + n)^2 = d^2 + 2 n d + n^2

p(both defective) = .12^2

Answer 3

It will be accepted
if both are good, OR one is good, and one is bad
= C(2,0)(.88)^2 + C(2,1)(.12)(.88)
= .7744 + .2112 = .9856

R_scott had:
p(both defective) = .12^2 = .0144

so prob(accepted) = 1- .0144 = .9856 , the same as my answer.