A bag contains a total of 8 batteries of which 3 is defective.
Selecting 3 at random without replacement, determine the probability that none of the batteries you select are good.
First we find the numerator of the probability:
We have 3 good batteries and we must choose all 3.
That's "3 choose 3". So the numerator is 3C3 or 1.
Next we find the denominator of the probability:
We have 8 batteries and we must choose any 3.
That's "8 choose 3". So the denominator is 8C3 or 56.
P(choosing all 3 bad batteries) =
Edwin