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An urn 996 black and 4 white balls. Drawing 50 randomly chosen balls without replacement what is the probability that there are all black?

My solution: $\prod_{i=0}^{49} (1 - \frac{4}{1000-i} ) = 0.814 $. In the brackets is the probability that a white ball wasn't selected. Each time we remove one balls, therefore the denominator $1000-i$. I must be making a mistake because the official answer is $0.7696$.

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  • $\begingroup$ Can't say anything about the calculation, but your answer is probably right. $\endgroup$ – Nikunj Mar 20 '16 at 11:56
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You're right, though your solution is unnecessarily complicated.

The probability is

$$ \frac{\binom{996}{50}}{\binom{1000}{50}}=\frac{950\cdot949\cdot948\cdot947}{1000\cdot999\cdot998\cdot997}\approx0.814\;. $$

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  • $\begingroup$ Joriki the hypergeometric distribution gives the probability that you just calculated but if we try binomia distribution..with $$p=\frac {996}{1000}$$ and $$q=\frac {4}{1000}$$ and $$n=50$$ we get the answer as 0.8184 roughyly..both the answers are approx equal but which is better..?? $\endgroup$ – Upstart Mar 20 '16 at 12:27
  • $\begingroup$ @Upstart: It's not a question of better or worse but of right and wrong :-) The binomial distribution is for drawing with replacement; here we're drawing without replacement. $\endgroup$ – joriki Mar 20 '16 at 12:28
  • $\begingroup$ Ohk thanx..idiot of me to ask this $\endgroup$ – Upstart Mar 20 '16 at 12:31

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