Question 1189346: * Show your complete solution
* Explain your answer after the solution
1. A shipment of 10 TV sets contains 3 defective units. If three units are taken for inspection, then what is the probability that:
a. all the defective TV sets are included
b. no defective TV set shall be included
c. only one of the defective TV sets shall be included
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 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
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