Ashipment of 12 microwave ovens contains three defective units. a vending company has ordered four units, and because each has identical packaging, the selection will be random. what is the probability that (a) all four units are good, (b) exactly two units are good, and (c) at least two units are good?

a) 54/55

b) 100/110

c) 99/110

Step-by-step explanation:

a)Probability of ordering 4= 9/12×3/11×2/10×1/9 = 54/11880

Probability of 4 good units= 4 × 54/11880

= 216/11880

1/55

1-(1/55) = 54/55

b)Probability of 2 good units= 2 × 54/(11880)

= 108/11880

= 1/110

1- (1/110)= 100/110

c) Probability of exactly 2 units not good= 1 -(100/110) =99/110

answer: 1,380

step-by-step explanation:

1: you turn %15 into a decimal by moving the decimal point two places to the left.

2: and then you multiply .15 by 1,200 , which will get you 180.00

3: from there you will add 180 to 1,200

4: which will get you 1,380

hope it *im kind of slow so idk if my answer is right*

3/6 chance

step-by-step explanation:

half of the nubers are odd numbers (1,3,5) which in that case if you would be told to simplify it would be 1/2