Answer:
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
