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CHAPTER 4 Special Probability Distributions

where, in the last two steps, we have respectively used the facts that the Xk are independent and are identically distributed. Now, by a Taylor series expansion, E[et(X1

m)>s1n]

cd s !n t2 t c E(X1 m) E[(X1 m)2] E(1) 2s2n s !n t t2 t2 c 1 c 1 (0) (s2) 2n 2s2n s !n t(X1 m) t2(X1 m)2 2s2n

* E(etSn)

E c1

so that

t2 2n

But the limit of this as n S ` is et 2>2, which is the moment generating function of the standardized normal distribution. Hence, by Theorem 3-10, page 80, the required result follows.

Multinomial distribution 4.26. A box contains 5 red balls, 4 white balls, and 3 blue balls. A ball is selected at random from the box, its color is noted, and then the ball is replaced. Find the probability that out of 6 balls selected in this manner, 3 are red, 2 are white, and 1 is blue.

Method 1 (by formula) P(red at any drawing) 5 12 P(white at any drawing) 3 12 625 5184 4 12

P(blue at any drawing) Then P(3 red, 2 white, 1 blue)

5 3 4 2 3 1 6! a b a b a b 3!2!1! 12 12 12

Method 2 The probability of choosing any red ball is 5>12. Then the probability of choosing 3 red balls is (5 >12)3. Similarly, the probability of choosing 2 white balls is (4 >12)2, and of choosing 1 blue ball, (3 >12)1. Therefore, the probability of choosing 3 red, 2 white, and 1 blue in that order is a 5 3 4 2 3 1 b a b a b 12 12 12

But the same selection can be achieved in various other orders, and the number of these different ways is 6! 3!2!1! as shown in 1. Then the required probability is 6! 5 3 4 2 3 1 a b a b a b 3!2!1! 12 12 12 Method 3 The required probability is the term p3 p2 pb in the multinomial expansion of (pr r w pw 4>12, pb 3>12. By actual expansion, the above result is obtained. pw pb)6 where pr 5>12,

The hypergeometric distribution 4.27. A box contains 6 blue marbles and 4 red marbles. An experiment is performed in which a marble is chosen at random and its color observed, but the marble is not replaced. Find the probability that after 5 trials of the experiment, 3 blue marbles will have been chosen.

Method 1 6 The number of different ways of selecting 3 blue marbles out of 6 blue marbles is a b . The number of different 3 4 ways of selecting the remaining 2 marbles out of the 4 red marbles is a b . Therefore, the number of different 2 6 4 samples containing 3 blue marbles and 2 red marbles is a b a b . 3 2

CHAPTER 4 Special Probability Distributions

4) in the box

Now the total number of different ways of selecting 5 marbles out of the 10 marbles (6 10 is a b . Therefore, the required probability is given by 5 6 4 a ba b 3 2 10 21 10 a b 5 Method 2 (using formula) We have b 6, r 4, n 5, x 3. Then by (19), page 113, the required probability is 6 4 a ba b 3 2 10 a b 2

The uniform distribution 4.28. Show that the mean and variance of the uniform distribution (page 113) are given respectively by 1 1 (a) m 2 (a b), (b) s2 12 (b a)2.

(a) (b) We have E(X2)