In a box of 20 bulbs, 2 are defective. What is the probability of choosing 2 bulbs that are not defective ?
I solved it by hypergeometric theorem and got it correct.
I want to know that how binomial theorem can solve this ?
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You cannot use the binomial distribution (theorem) because it is without replacement. – callculus42 Nov 30 '16 at 19:29
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@callculus, plz elaborate ? – Jon Garrick Nov 30 '16 at 19:31
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1One assumption which has to be made to use the binomial distribution is that you have an experiment without replacement. – callculus42 Nov 30 '16 at 19:34
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@callculus, Then how the question will be changed inorder to apply BT ? – Jon Garrick Nov 30 '16 at 19:39
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1Every bulb which has been drawn has to be replaced before the next bulb is drawn. But then it is another setting/exercise. – callculus42 Nov 30 '16 at 19:42
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Maybe of interest http://math.stackexchange.com/questions/330553/proof-that-the-hypergeometric-distribution-with-large-n-approaches-the-binomia or http://wiki.stat.ucla.edu/socr/index.php/AP_Statistics_Curriculum_2007_Limits_Bin2HyperG – cgiovanardi Dec 01 '16 at 00:02
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edit: The binomial theorem relies on the probabilities of success and failure being constant, whereas here they are not. Once you remove the lightbulb, success has a new probability of $17/19$, no longer $18/20$, as it would be with replacement.
What is the probability of the first bulb not being defective? $18/20=9/10$.
And the second? $17/19$. You have removed one of the good lightbulbs, so you must subtract 1 from the original numerator as well as the denominator.
This yields a probability of both being good bulbs $$ \frac{9}{10}*\frac{17}{19}=\frac{153}{190} $$
operatorerror
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1The question is "I want to know that how binomial theorem can solve this " I don´t see that you answer referring to that. – callculus42 Nov 30 '16 at 19:36
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It cannot, as you commented. I offered an intuitive way to solve the problem – operatorerror Nov 30 '16 at 19:37
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1I like this kind of intuitive ways, much better than just using the hypergeometric distribution. But it doesn´t answer the question. – callculus42 Nov 30 '16 at 19:39
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1@callculus in that case, the answer is just "you cannot use the binomial theorem to solve this." Would you be satisfied if I added this in at the start? – operatorerror Nov 30 '16 at 19:41
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@qbert, yes please !! I need the reasoning. Actually I solved it by this method too. – Jon Garrick Nov 30 '16 at 19:44
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Binomial coefficients come in handy here.
$$\frac {\binom {18}2}{\binom {20}2}=\frac {153}{190}$$
Hypergeometricx
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