Page 37 - Probability, Random Variables and Random Processes

CHAP. 1) PROBABILITY

Show that for any events A and B in S,
P(B) = P(B I A)P( A) + P(B I A) P(X)

From Eq. (1.64) (Prob. 1.23), we have
P(B) = P(B n A) + P(B n 4

Using Eq. (1.39), we obtain
P(B) = P(B I A)P(A) + P(B I X)P(A)
Note that Eq. (1.83) is the special case of Eq. (1.44).

Suppose that a laboratory test to detect a certain disease has the following statistics. Let
A = event that the tested person has the disease
B = event that the test result is positive

It is known that
P(B I A) = 0.99 and P(B I A) = 0.005

and 0.1 percent of the population actually has the disease. What is the probability that a person
has the disease given that the test result is positive?
From the given statistics, we have

P(A) = 0.001 then P(A) = 0.999
The desired probability is P(A ) B). Thus, using Eqs. (1.42) and (1.83), we obtain

Note that in only 16.5 percent of the cases where the tests are positive will the person actually have the
disease even though the test is 99 percent effective in detecting the disease when it is, in fact, present.

A company producing electric relays has three manufacturing plants producing 50, 30, and 20
percent, respectively, of its product. Suppose that the probabilities that a relay manufactured by
these plants is defective are 0.02,0.05, and 0.01, respectively.

If a relay is selected at random from the output of the company, what is the probability that
it is defective?
If a relay selected at random is found to be defective, what is the probability that it was
manufactured by plant 2?

Let B be the event that the relay is defective, and let Ai be the event that the relay is manufactured by
plant i (i = 1,2, 3). The desired probability is P(B). Using Eq. (1.44), we have