Question 1189346
The chance of the first TV being defective is 3/10. Now there are 9 left of which 2 are defective, so the probability is 2/9. Then 1/8.
In the second it is 7/10 for the first, 6/9 for the second, and 5/8 for the third.
With only one of the TVs, it can happen with the first, second, or third with equal probability, so you need a 3. Then multiply that 3 by one possibility, such as the first one (3/10) then 7/9, not defective with 7 of them and 9 left, and 6/8.


a. 3/10*(2/9)(1/8)=6/720=1/120
b. (7/10)(6/9)(5/8)=210/720 or 7/24
c. 3*(3/10)(7/9)(6/8)=378/720=21/40; three ways to have the defective TV included
This is also 3C1*7C2/10C3=3*21/120=21/40