Find The Probability

Question

Defective units a shipment of 12 microwave ovens contains three defective units. A vending company has ordered four of these units, and because each is identically packaged, the selection will be random. What are the probabilities that (a) all four units are good, (b) exactly two units are good, and (c) at least two units are good?

Answer 1

12 ovens, 3 defective, 9 not defective

pick 4 at random,

probability all are not defective = 4C4(9/12)(8/11)(7/10)(6/9)= (3/4)(8/11)(7/10)(2/3)

= (6/11)(7/15) = 14/55 = .25454545... = about 25.5% probability of all 4 not defective

probability of exactly 2 not defective, and exactly 2 defective = 4C2(9/12)(8/11)(3/10)(2/9)

= 4!/2!2!(3/4)(8/11)(1/15) = 6(6/11)(1/15) = 12/55 = .2181818... = about 21.8%

probability of at least 2 good = P(4 good) + P(3 good) + P(2 good) = 14/55 + P(3 good) + 12/55

= 26/55 + P(3 good) = 26/55 + 4C3(9/12)(8/11)(7/10)(3/9) = 26/55 + 28/55 = 54/55 = about 98.2%

probability of at least 2 good = One - Probability of zero good - probability of exactly one good

= 1 - 0 - 4C3(3/12)(2/11)(1/10)(9/9) = 1 - 4(3/660) = 1-4/220 = 1-1/55 = 54/55 = 98.2%