Asked by Sara!
in a shipment of 20 computers, 3 are defective
three computers are randomly selected and tested. what is the probability that all 3 are defective if the first and second ones are not replaced after being tested.
a. 1/760
b.1/1140
c.27/8000
d.3/5000
i chose D, is this correct
three computers are randomly selected and tested. what is the probability that all 3 are defective if the first and second ones are not replaced after being tested.
a. 1/760
b.1/1140
c.27/8000
d.3/5000
i chose D, is this correct
Answers
Answered by
Damon
First one 3/20
Second one 2/19
third one 1/18
3/20*2/19*1/18 = 1/(3*20*19) =1/1140
Second one 2/19
third one 1/18
3/20*2/19*1/18 = 1/(3*20*19) =1/1140
Answered by
Hope
The answer is B Explanation: 3/20*2/19*1/18=1/(3*20*19)=1/1140
Answered by
Srishti Sah
In a shipment of 20 computers, 3 are defective. Two computers are randomly selected and tested. What is the probability that both are defective if the first and second ones are not replaced after being tested?
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