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
Answer is .491 but I can not figure out how they come up with that!
Solution
P(None defective) = 16C3/20C3
(As there are 16 non defective alternators and 20 total alternators
=> Probability = ways of selecting 3 alternatorsout of 16 / ways of selecting 3 alternators out of
20)
= 0.491.
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A batch of 20 used automobile alternators contains 4 defectives. If .pdf
1. 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
Answer is .491 but I can not figure out how they come up with that!
Solution
P(None defective) = 16C3/20C3
(As there are 16 non defective alternators and 20 total alternators
=> Probability = ways of selecting 3 alternatorsout of 16 / ways of selecting 3 alternators out of
20)
= 0.491
