SOLUTION: suppose you receive a shipment of 30 computers, 2 of which are defective. What is the probability of picking a sample of 5 computers with 2 defective computers in it?

Question 619294: suppose you receive a shipment of 30 computers, 2 of which are defective. What is the probability of picking a sample of 5 computers with 2 defective computers in it?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
probability of getting exactly 2 defectives in a sample of 5 from a batch of 30 with 2 defectives in it can be calculated on 2 ways.
the first way:
the probability of first draw being defective is 2/30
assuming the first draw is successful, the probability of second draw being defective is 1/29
assuming the second draw is successful, the probability of third draw not being defective is 28/28
assuming the third draw is successful, the probability of fourth draw not being defective is 27/27
assuming the fourth draw is successful, the probability of fifth draw not being defective is 26/26
number of ways you can get 2 defectives out of 5 is equal to C(5,2) which is equal to 10
probability is:
2/30 * 1/29 * 28/28 * 27/27 * 26/26 * 10 which is equal to:
.022988506
the second way:
number of possible ways you can get 2 defectives out of 2 is equal to C(2,2) which is equal to 1.
number of possible ways you can get 3 non defectives out of 28 is equal to C(28,3) which is equal to 3276.
number of possible ways you can get 5 total out of 30 is equal to C(30,5) whihc is equal to 142506.
probability is:
(1 * 3276) / 142506 which is equal to:
.022988506
get the same answer either way which is good.
C(n,x) is the combination formula for getting x out of a set of n.
the formula is:
C(n,x) = n! / (x! * (n-x)!)
Example:
C(30,5) = 30! / (5! * 25!) which is equal to 2.652528598*10^32 / (120 * 1.551121004*10^25), the whole thing being equal to 142506.
you need your calculator to confirm this is true.

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