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

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

Answer 2

P(defective) = 3/12 = 1/4
P(good) = 1 - 1/4 = 3/4
$P(2\ good)=4C2\times0.75^{2}\times0.25^{2}=0.211$
$P(3\ good)=4C3\times0.75^{3}\times0.25=0.422$
$P(4\ good)=0.75^{4}=0.316$
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.