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5-2 MULTIPLE DISCRETE RANDOM VARIABLES 155

The multinomial distribution is considered a multivariable extension of the binomial
distribution.

EXAMPLE 5-13 In Example 5-12, let the random variables X 1 , X 2 , X 3 , and X 4 denote the number of bits that are
E, G, F, and P, respectively, in a transmission of 20 bits. The probability that 12 of the bits
received are E, 6 are G, 2 are F, and 0 are P is

20!
6
2
0
12
12, X 6, X 2, X 02 0.6 0.3 0.08 0.02 0.0358
P1X 1 2 3 4
12!6!2!0!
Each trial in a multinomial random experiment can be regarded as either generating or not
generating a result in class i, for each i 1, 2, . . . , k. Because the random variable X i is the
number of trials that result in class i, X i has a binomial distribution.

If X , X , . . . , X have a multinomial distribution, the marginal probability distribu-
2
k
1
tion of X is binomial with
i
E1X i 2 np i and V 1X i 2 np i 11 p i 2 (5-14)

EXAMPLE 5-14 In Example 5-13, the marginal probability distribution of X is binomial with n 20 and
2
p 0.3. Furthermore, the joint marginal probability distribution of X and X is found as
3
2
follows. The P(X x , X x ) is the probability that exactly x trials result in G and that x 3
2
2
3
3
2
result in F. The remaining n x x trials must result in either E or P. Consequently, we can
3
2
consider each trial in the experiment to result in one of three classes, {G}, {F}, or {E, P}, with
probabilities 0.3, 0.08, and 0.6 0.02 0.62, respectively. With these new classes, we can
consider the trials to comprise a new multinomial experiment. Therefore,
f 1x , x 2 P1X x , X x 2
3
3
2
2
2
3
X 2 X 3
n!
10.32 10.082 10.622 n x 2 x 3
x 2
x 3
x !x ! 1n x x 2!
2
3
2
3
The joint probability distribution of other sets of variables can be found similarly.
EXERCISES FOR SECTION 5-2
5-17. Suppose the random variables X, Y, and Z have the Determine the following:
following joint probability distribution (a) P 1X 22 (b) P 1X 1, Y 22
(c) P 1Z 1.52 (d) P 1X 1 or Z 22
x y z f(x, y, z) (e) E 1X 2
1 1 1 0.05 5-18. Continuation of Exercise 5-17. Determine the follow-
1 1 2 0.10 ing:
(a) P 1X 1 ƒ Y 12 ( b ) P 1X 1, Y 1 ƒ Z 22
1 2 1 0.15
(c) P 1X 1 ƒ Y 1, Z 22
1 2 2 0.20
5-19. Continuation of Exercise 5-17. Determine the condi-
2 1 1 0.20
tional probability distribution of X given that Y 1 and Z 2.
2 1 2 0.15
5-20. Based on the number of voids, a ferrite slab is classi-
2 2 1 0.10
fied as either high, medium, or low. Historically, 5% of the
2 2 2 0.05
slabs are classified as high, 85% as medium, and 10% as low.