A) There are 20 machines in a factory. 7 of the machines are defective.
Assuming you randomly choose one, place it back in the pile and randomly choose the second one, what is the probability that both choices will be defective machines?
B) There are 20 machines in a factory. 7 of the machines are defective.
Assuming you randomly choose one, do not place it back in the pile and randomly choose the second one, what is the probability that both choices will be defective machines?
A) Total machine = 20
Defective machine =7
Non defective machine = 13
P(1st machine being defective)= 7/20
P(2nd machine being defective when first machine is placed back ) = 7/20
So PROBABILITY that both the machine being defective is = (7/20)×(7/20) = 49/400
So answer is 49/400
B)
P(1st machine being defective) =7/20
P(2nd being defective when first machine is not returned/replaced) = 6/19
So PROBABILITY that both the machine being defective (when not replaced) = (7/20)×(6/19)
=42/380
=21/190
So answer is 21/190
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