B. P. Lathi, Zhi Ding - Modern Digital and Analog Communication Systems-Oxford University Press (2009)

8.1 Concept of Probability 401
But the event of "k successes in n trials" can occur in many different ways (different orders).
It is well known from combinatorial analysis that there are
( n )
k
n!
- (8.16)
k!(n - k)!
ways in which k positions can be taken from n positions (which is the same as the number of
ways of achieving k successes in n trials).
This can be proved as follows. Consider an urn containing n distinguishable balls marked
1, 2, ... , n. Suppose we draw k balls from this urn without replacing them. The first ball could
be any one of the n balls, the second ball could be any one of the remaining (n - 1) balls, and
so on. Hence, the total number of ways in which k balls can be drawn is
n!
n(n - l)(n - 2) ... (n - k + 1) = --­
(n - k)!
Next, consider any one set of the k balls drawn. These balls can be ordered in different ways.
We could label any one of the k balls as number 1, and any one of the remaining (k - 1) balls
as number 2, and so on. This will give a total of k(k - l)(k - 2) • • • 1 = k! distinguishable
patterns formed from the k balls. The total number of ways in which k things can be taken
from n things is n!/(n - k)! But many of these ways will use the same k things, arranged in
different order. The ways in which k things can be taken from n things without regard to order
(unordered subset k taken from n things) is n!/ (n - k) ! divided by k! This is precisely defined
by Eq. (8.16).
This means the probability of k successes in n trials is
P(k successes in n trials) = G) p k (1 - p r - k
n! k (l )
n-k
k!(n - k)! p - p (8.17)
Tossing a coin and observing the number of heads is a Bernoulli trial with p = 0.5. Hence, the
probability of observing k heads in n tosses is
P(k heads in n tosses) =
(0.5) k (0.5t - k
( n n!
)
= ---(O.St
k
k!(n - k)!
Exam pie 8. 6
A binary symmetric channel (BSC) has an error probability P e (i.e., the probability of receiving
0 when 1 is transmitted, or vice versa, is P e )- Note that the channel behavior is symmetrical
with respect to O and 1. Thus,
P(Ol l) = P(l lO) = P e
and
P(OIO) = P(l i l) = 1 - P e
where P(y lx) denotes the probability of receiving y when x is transmitted. A sequence of n
binary digits is transmitted over this channel. Determine the probability of receiving exactly
k digits in error.