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5-1 TWO DISCRETE RANDOM VARIABLES 149

For discrete random variables X and Y, if any one of the following properties is true,
the others are also true, and X and Y are independent.
(1) f XY 1x, y2 f 1x2 f 1 y2 for all x and y
Y
X
(2) f 1 y2 f 1y2 for all x and y with f X 1x2
0
Y
Y ƒ x
(3) f 1x2 f 1x2 for all x and y with f 1 y2
0

Y
X ƒy X
(4) P1X A, Y B2 P1X A2 P1Y B2 for any sets A and B in the range
of X and Y, respectively. (5-7)

Rectangular Range for (X, Y)!
If the set of points in two-dimensional space that receive positive probability under
f (x, y) does not form a rectangle, X and Y are not independent because knowledge of X
XY
can restrict the range of values of Y that receive positive probability. In Example 5-1
knowledge that X 3 implies that Y can equal only 0 or 1. Consequently, the marginal
1y2
probability distribution of Y does not equal the conditional probability distribution f Y 03
for X 3. Using this idea, we know immediately that the random variables X and Y with
joint probability mass function in Fig. 5-1 are not independent. If the set of points in two-
dimensional space that receives positive probability under f (x, y) forms a rectangle,
XY
independence is possible but not demonstrated. One of the conditions in Equation 5-7 must
still be verified.
Rather than verifying independence from a joint probability distribution, knowledge of
the random experiment is often used to assume that two random variables are independent.
Then, the joint probability mass function of X and Y is computed from the product of the
marginal probability mass functions.

EXAMPLE 5-9 In a large shipment of parts, 1% of the parts do not conform to specifications. The supplier
inspects a random sample of 30 parts, and the random variable X denotes the number of parts
in the sample that do not conform to specifications. The purchaser inspects another random
sample of 20 parts, and the random variable Y denotes the number of parts in this sample that
do not conform to specifications. What is the probability that X 1 and Y 1 ?
Although the samples are typically selected without replacement, if the shipment is large,
relative to the sample sizes being used, approximate probabilities can be computed by assum-
ing the sampling is with replacement and that X and Y are independent. With this assumption,
the marginal probability distribution of X is binomial with n 30 and p 0.01, and the mar-
ginal probability distribution of Y is binomial with n 20 and p 0.01.
If independence between X and Y were not assumed, the solution would have to proceed
as follows:

P1X 1, Y 12 P1X 0, Y 02 P1X 1, Y 02
P1X 0, Y 12 P1X 1, Y 12
f 10, 02 f XY 11, 02 f XY 10, 12 f XY 11, 12
XY
However, with independence, property (4) of Equation 5-7 can be used as

P1X 1, Y 12 P1X 12 P1Y 12