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?
Do i have to work with combination and Laplace?
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
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.
To calculate the probability that the supplier will be accepted, we need to determine the probability that both examined goods meet the requirements.
First, let's calculate the probability that the first examined good meets the requirements.
Out of the 25 delivered items, 3 are defective. So, the probability of randomly selecting a non-defective item on the first try is (25 - 3) / 25 = 22 / 25.
Next, let's calculate the probability that the second examined good meets the requirements.
After the first item is examined, there are 24 items left, with 2 defective ones. So, the probability of randomly selecting a non-defective item on the second try is (24 - 2) / 24 = 22 / 24.
To calculate the probability that both examined goods meet the requirements, we multiply the probability of the first examined good meeting the requirements by the probability of the second examined good meeting the requirements:
P(both goods meet requirements) = (22/25) * (22/24)
P(both goods meet requirements) = 0.88 * 0.9167
P(both goods meet requirements) ≈ 0.805
Therefore, the probability that the supplier will be accepted is approximately 0.805, or 80.5%.