A batch of 20 used automobile alternators contains 4 defectives. If 3 alternators are selected at ra

A batch of 20 used automobile alternators contains 4 defectives. If 3 alternators are selected at random, find the probability of the events (Using the Rule of combinations):
A= None of the defective appears
B= Exactly two defective appears
The answer for A is .491 the answer for B is .084 but I can not figure out for the life of me how they got these answers.
I’ve tried 3 of 20 which gave the 1140 total selections possible. Than if there is 4 defective and I want the probability of selecting none. I tried 0 of 4 which doesn’t work 1 of 4 likewise doesn’t work. I’m stuck. Please help I know this is probably going to be something so simple I am missing.

 

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