A shipment of 12 television sets contains 3 defective sets. In how many ways can a hotel purchase 5 of these sets and receive at least 2 of the defective sets?

Question

Please help

Answer 1

Two separate sets: 9 “good” TVs and 3 “bad” TVs make up 12 TV sets.

The 5 TVs delivered to the hotel must consist of at least 2 bad TVs. So that could be 3 bad TVs and 2 good ones; or 3 good TVs and 2 bad ones.

There is only one way to get 3 bad TVs, because there are only 3 of them anyway. But there are 9×8/2=36 ways of getting 2 good TVs out of 9.

There are 3 ways of getting 2 bad TVs out of 3, and 9×8×7/(2×3)=84 ways of getting 3 good TVs out of 9. Combine these combinations: 3×84=252.

Add these combinations together (2 bad/3 good+2 good/3 bad): 252+36=288 ways of getting at least 2 defective TVs.

(If this were to be converted to a probability, we’d divide 288 by the number of ways 5 TVs could be selected out of 12, 288/792=4/11.)

Answer 2

Solution:

N= 3C2 * 9C3 + 3C3 * 9C2

N= 288 ways