Question 368260: Quality Control: A factory produces fuses, which are packaged in boxes of 10. Three fuses are selected at random from each box for inspection. The box is rejected if at least one of these three fuses is defective. What is the probability that a box containing five defective fuses will be rejected?
The answer is 11/12 which = .917, but how is it done?
Thank you in advanced...
Found 2 solutions by robertb, sudhanshu_kmr:
Answer by robertb(5830)   (Show Source):
You can put this solution on YOUR website! The experiment follows a hypergeometric distribution: there are 5 good fuses and 5 bad fuses.
1st case, 1 bad and 2 good fuses: # of ways is C(5,1)*C(5,2) = 50.
2nd case, 2 bad and 1 good fuse: # of ways is C(5,2)*C(5,1) = 50.
3rd case, 3 bad and 0 good fuse: # of ways is C(5,3)*C(5,0) = 10.
The number of ways of choosing any 3 out of 10 fuses in a box is C(10,3) = 120.
Therefore the probability is (50+50+10)/120 = 110/120 = 11/12.
Answer by sudhanshu_kmr(1152)  (Show Source):
You can put this solution on YOUR website! First we find the probability that in all three consecutive test
fuses are not defective.
probability of first fuse is not defective = 5/10 (in total 10, 5 no defetive)
as now there are 4 non-defective fuse remain in total 9.
probability of second fuse is not defective = 4/9 (in total 9, 4 non defective)
probability of third fuse is not defective = 3/8 (in total 8, 3 non defective)
thus probability = 5/10 * 4 /9 * 3/8 = 1/12
we know p' = 1-p
thus the probability of not happening it = 1 - 1/12 = 11/12
probability that at least one of 3 test will fail = 11/12
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