Question 495961
probability of an item being defective is .1
probability of an item not being defective is 1 - .1 = .9
if p = defective and q = not defective, then the formula for x out of n would be:
p(x) = p^x * q^(n-x) * nCx
we are drawing 3 items from the batch, so n = 3
we want to find the probability that at least 2 are defective out of the 3.
this would be equal to 1 - p(0) - p(1) or it would be equal to p(2) + p(3).
we'll work it both ways so you can see how that happens.
nCx is the combination formula for x things out of n.
it is equal to n! / (x! * (n-x)!)
using the formula of p(x), we get:
p(0) = .1^0 * .9^3 * 3C0 = .729 * 1 = .729
p(1) = .1^1 * .9^2 * 3C1 = .081 * 3 = .243
p(2) = .1^2 * .9^1 * 3C2 = .009 * 3 = .027
p(3) = .1^3 * .9^0 * 3C3 = .001 * 1 = .001
sum up all the probabilities and you get p(all) = 1
this is what you should get when you add up all the probabilities of independent events like we have here.
p(>=2) = p(2) + p(3) = .027 + .001 = .028
p(>=2) = 1 - p(0) - p(1) = 1 - .729 - .243 = .028
we get the same answer either way, as we should.
your answer is that the probability of getting a sample with at least 2 defective units in it is .028 or 2.8%