math
posted by gift onos .
A box contain 12 cell phones of which 4 are defective.all phones look alike.3 phones are drawn randomly what is the probability.1 all the 3 bulbs are defective.2 at least 2 of the bulbs chosen are defective.3 at most 2 of the bulbs chosen are defective.hint use combination analysis

prob(defective) = 4/12 = 1/3
prob(not defective) = 2/3
1. all 3 defective  prob = (1/3)^3 = 1/27
2. at least 2 defective
> 2 defective or 3 defective
prob = C(3,2)(1/3)^2 (2/3) + (1/3)^3
= 3(2/27) + 1/27 = 7/27
3. at most 2 > 0 defective, 1 defect, or 2 defective
or exclude all 3 defectiv
= 1  1/27 = 26/27
BTW,
Where do the bulbs enter the picture when talking about cell phones