Ashipment 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

look a the sides they are perpendicular

answer: ur v dumb

step-by-step explanation: