Question

A shipment of 12 microwave ovens contains 3 defective units. A vending company has ordered 4 of these units, and because all are packaged identically, the selection will be at random. What is the probability that at least 2 units are good?

Answer 1

P(defective) = 3/12 = 1/4
P(good) = 1 - 1/4 = 3/4

`P(2\ good)=4C2*0.75^(2)*0.25^(2)=0.211`

`P(3\ good)=4C3*0.75^(3)*0.25=0.422`

`P(4\ good)=0.75^(4)=0.316`
The probability that at least 2 units are good is given by:
P(2 good) + P(3 good) + P(4 good) = 0.211 + 0.422 + 0.316 = 0.949.

Answer 2

In this problem, 4 units must be chose from a shipment of 12 microwave ovens containing 3 defective units.

To solve for the probability that at least 2 units are good,
[(9c2 x 3c2) + (9c3 x 3c1) + 9c4] / 12c4
= 486/495
= 54/55 or 0.9818