Page 37 - Schaum's Outlines - Probability, Random Variables And Random Processes

PROBABILITY [CHAP 1

(b) The desired probability is P(A2 1 B). Using Eq. (1.42) and the result from part (a), we obtain

1.51. Two numbers are chosen at random from among the numbers 1 to 10 without replacement. Find
the probability that the second number chosen is 5.
Let A,, i = 1, 2, . . . , 10 denote the event that the first number chosen is i. Let B be the event that the
second number chosen is 5. Then by Eq. (1.44),

Now P(A,) = A. P(B I A,) is the probability that the second number chosen is 5, given that the first is i. If
i = 5, then P(B I Ai) = 0. If i # 5, then P(B I A,) = 4. Hence,

1.52. Consider the binary communication channel shown in Fig. 1-15. The channel input symbol X
may assume the state 0 or the state 1, and, similarly, the channel output symbol Y may assume
either the state 0 or the state 1. Because of the channel noise, an input 0 may convert to an
output 1 and vice versa. The channel is characterized by the channel transition probabilities p,,
40, PI, and 91, ckfined by

where x, and x, denote the events (X = 0) and (X = I), respectively, and yo and y, denote the
events (Y = 0) and (Y = I), respectively. Note that p, + q, = 1 = p, + q,. Let P(xo) = 0.5, po =
0.1, and p, = 0.2.

(a) Find P(yo) and P(yl).
(b) If a 0 was observed at the output, what is the probability that a 0 was the input state?
(c) If a 1 was observed at the output, what is the probability that a 1 was the input state?
(d) Calculate the probability of error P,.

Fig. 1-15

(a) We note that