Page 38 - Schaum's Outlines - Probability, Random Variables And Random Processes

CHAP. 11 PROBABILITY

Using Eq. (1.44), we obtain

(b) Using Bayes' rule (1.42), we have

(c) Similarly,

(d) The probability of error is
P, = P(yl (xo)P(xo) + P(yo (xl)P(xl) = O.l(O.5) + 0.2(0.5) = 0.15.

INDEPENDENT EVENTS
1.53. Let A and B be events in a sample space S. Show that if A and B are independent, then so are (a)
A and B, (b) A and B, and (c) A and B.

(a) From Eq. (1.64) (Prob. 1.23), we have
P(A) = P(A n B) + P(A n B)
Since A and B are independent, using Eqs. (1.46) and (1 .B), we obtain
P(A n B) = P(A) - P(A n B) = P(A) - P(A)P(B)
= P(A)[l - P(B)] = P(A)P(B)
Thus, by definition (l.46), A and B are independent.
(b) Interchanging A and B in Eq. (1.84), we obtain
P(B n 3 = P(B)P(A)
which indicates that A and B are independent.
(c) We have
P(A n B) = P[(A u B)] [Eq. (1.1411
= 1 - P(A u B) [Eq- (1.25)1
= 1 - P(A) - P(B) + P(A n B) [Eq. (1.29)]
= 1 - P(A) - P(B) + P(A)P(B) [Eq. (1.46)]
= 1 - P(A) - P(B)[l - P(A)]
= [l - P(A)][l - P(B)]
= P(A)P(B) [Eq. (1.2511
Hence, A and B are independent.

1.54. Let A and B be events defined in a sample space S. Show that if both P(A) and P(B) are nonzero,
then events A and B cannot be both mutually exclusive and independent.
Let A and B be mutually exclusive events and P(A) # 01, P(B) # 0. Then P(A n B) = P(%) = 0 but
P(A)P(B) # 0. Since

A and B cannot be independent.

1.55. Show that if three events A, B, and C are independent, then A and (B u C) are independent.