elemprob-fall2010-page18

elemprob-fall2010-page18 - 5 2 3 2 8 4 . If there are m of...

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we are looking at the number of successes in 365 “trials.” In other words, we are looking at a binomial. The above proposition says that this can be approximated very well by a Poisson. If instead of days, we look at minutes, the approximation is even better. 7 Hypergeometric distributions Suppose we have an urn with 5 red marbles and 3 green marbles. We pick 4 marbles without replacement and ask, what is the probability we have 2 red marbles and 2 green marbles? We could also pick 4 marbles with replacement and get a different answer. There are ± 8 4 ² ways of choosing 4 marbles out of 8 total, all equally likely. There are ± 5 2 ² ways of choosing 2 red marbles out of the 5 red ones and ± 3 2 ² ways of choosing the 2 green, so altogether the probability is
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Unformatted text preview: 5 2 3 2 8 4 . If there are m of one type, n of another, and we want N total to be chosen, with k of hte rst type, the probability becomes m k n N-k m + n N . A random variable with these probabilities ( m,n,N are parameters, k the possible values) is said to have a hypergeometric distribution. An example. What is the probability that in a poker hand (5 cards out of 52) we get exactly 4 of a kind? Answer: The probability of 4 aces and 1 king is 4 4 4 1 . 52 5 . The probability of 4 jacks and one 3 is the same. There are 13 ways to pick the 18...
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