Question 1178926: A shipment of 10 TV sets contains 2 units that are defective. In how many ways can 3 TV sets be inspected such that:
a. no defective TV set is included?
b. both defective TV sets are included for inspection?
c. only one of the defective TV sets is included?
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
a. no defective TV set is included --> choose 0 of the 2 bad TVs and 3 of the 8 good TVs: C(2,0)*C(8,3)
b. both defective TV sets are included for inspection --> choose 2 of the 2 bad TVs and 1 of the 8 good TVs: C(2,2)*C(8,1)
c. only one of the defective TV sets is included --> choose 1 of the 2 bad TVs and 2 of the 8 good TVs: C(2,1)*C(8,2)
If that is not enough of an answer for you, re-post the question telling us what it is about the problem that you are having trouble with.
|
|
|
