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Approximation to Binomial Distribution If N and K are very large with respect to n and k, then the hypergeometric distribution can be approximated by the binomial distribution as shown below: P{E ¼ k} ¼ K k NÀK nÀk 0 N n ¼ k! (N À K)! (N À n)! (K À k)! (N À K À n þ k)! N! ¼ n! K! (N À n)! (N À K)! (n À k)! (K À k)! N! (N À K À n þ k)! ¼ n K(K À 1) Á Á Á (K À k þ 1) k N(N À 1) Á Á Á (N À k þ 1) Â (N À K)(N À K À 1) Á Á Á (N À K À n þ k þ 1) (N À k)(N À K À 1) Á Á Á (N À k À n þ k þ 1) (4:6:3) Since K ) k and N ) n, Eq.

We need to find P{X} from the conditional probabilities P{X j A} ¼ 0:04, P{X j B} ¼ 0:02, P{X j C} ¼ 0:05, and P{X j D} ¼ 0:01. We are also given that P{A} ¼ 0:2, P{B} ¼ 0:3, P{C} ¼ 0:1, and P{D} ¼ 0:4. Substituting these quantities in Eq. 3%. Bayes’ Theorem The next logical question to ask is, “Given that the event B has occurred, what are the probabilities that A1, A2, . . ” In other words, given the conditional probabilities P{B j A1 }, P{B j A2 }, . . , P{B j An }, we have to find the reverse conditional probabilities P{A1 j B}, P{A2 j B}, .

3 Combinations 29 Sampling without Replacement and without Ordering We now discuss the case where we want to obtain a sample of r balls from n balls without replacement and without ordering. We have already seen that the number of ways to get r balls from n balls with ordering is (n)r. If we disregard ordering, then the number of ways is called the r-combination from the n balls. We can define this quantity as n {r-combination from n balls} ¼ (3:3:1) r ÀnÁ By ordering the elements in each r-combination, we can obtain from ÀnÁ r the rpermutations (n)r.