A shipment of 9 microwave ovens contains 2 defective units. A restaurant buys four of these units. What is the probability of the restaurant buying at least three nondefective units?

the probability of choosing more than 3 non defective units out of a sample of 4 is 0.784 (or 78.4%)

Step-by-step explanation:

Assuming that all the microwaves come from shipments selected at random, such that the probability of each microwave to be defective is independent from the others, then the random variable X = number of microwave units that are defective, has a binomial distribution such that the probability distribution is

P (X) = n!/[ (n-x) !*x!]*p^x * (1-p) ^ (n-x)

where

n = number of microwaves bought=4

x = number of microwaves that are defective

p = probability that a microwave has defective units = 2/9

then the probability that the restaurant buys at least 3 nondefective units is

Pfinal = P (X≤1) = P (X=1) + P (X=0) = 0.418 + 0.366 = 0.784