SOLUTION: Two out of six computers in a lab have problems with hard drives. If three computers are selected at random for inspection, what is the probability that none of them has hard drive

Question 1105404: Two out of six computers in a lab have problems with hard drives. If three computers are selected at random for inspection, what is the probability that none of them has hard drive problems
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
2 out of 6 have problems with hard drives.

this means that 4 out of 6 don't.

you choose 3 computers in a row from the batch of 6.

the probability that the first one chosen does not have a problem is 4/6 because there are 4 out of 6 in the batch that do not have a problem.

assuming the first one chosen does not have a problem and you choose again, then the probability that the second one chosen does not have a problem is 3/5 because there are now 3 out of 5 in the batch that do not have a problem.

assuming the second one chosen does not have a problem and you choose again, then the probability that the third one chosen does not have a problem is 2/4 because there are now 2 out of 4 in the batch that do not have a problem.

the probability that all 3 do not have a problem is therefore 4/6 * 3/5 * 2/4 = (4*3*2) / (6*5*4) = 24/120 = 12/60 = 6/30 = 3/15 = 1/5.

alternatively, you can use the basic definition of probability, which is the number of ways you can get a set of 3 non-defective components out of a set of 6 components, some of which are defective and some of this was not defective.

in this alternative, you would use the combination formula of c(n,x) = n! / (x! * (n-x)!)

use of this formula involves the following.

the number of ways you can get 3 non defective components out of 4 non-defective components is c(4,3) = 4! / (3! * 1!) = 4.

the number of ways you can get 3 components out of 6 components without regard to whether they are defective or not, is c(6,3) = 20.

the probability of getting 3 non-defective components out of the batch is therefore 4/20 which is equal to 1/5.

this is an example of probability without replacement, since, once your draw a non-defective component out of the batch, you do not put it back, in the batch, therefore changing the probability that the next component drawn out of the batch will not be defective.

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