Question 983868: Package of 10 game boy sets contains 3 defective sets. If 5 sets are to be picked out randomly and sent to a customer for an inspection, in how many ways can the customer find at least two defective set?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Look at the probability of 0 and 1, and the complement is at least 2.
Probability of 0 defective sets are (7/10)*(6/9)(5/8)(4/7)(3/6)=2520/30240=0.0833
Probability of one defective set (1 way)
(7/10)(6/9)(5/8)(3/7)(4/6). The denominator stays the same. The numerator changes its order but not the numbers. It is also 2520 (by chance)/30240. But there are 5 ways this can occur, with the defective first, second, third, or fifth being the other four.
That = 0.416667
The total of these two = 0.5
The complement is 0.5
For 2, if one chooses this approach.
(7/10)(6/9)(5/8)(3/7)(2/6)=1260/30240. There are 10 ways this can happen, 5C2, and the probability is 0.4166667
For 3, it is (3/10)(2/9)(1/8)(1)(1)=0.00833, but there are 10 ways this can happen as well, so the probability is 0.083333
These add up to 0.5, and that is a check.
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