Practice 5

Problem 7 (Conditional probability, Bayes` rule). Radar detection. If an aircraft is present in a certain area, a radar correctly registers its presence with probability 0.99. If it is not present, the radar falsely registers an aircraft presence with probability 0.10. We assume that an aircraft is present with probability 0.05. What is the probability of false alarm (a false indication of aircraft presence), and the probability of missed detection (nothing registers, even though an aircraft is present)?
A sequential representation of the sample space is appropriate here, as shown in Fig. 1.

Figure 1: Sequential description of the sample space for the radar detection problem

Solution:
Let A and B be the events

A={an aircraft is present}, B={the radar registers an aircraft presence},

and consider also their complements

Ac={an aircraft is not present}, Bc={the radar does not register an aircraft presence}.

The given probabilities are recorded along the corresponding branches of the tree describing the sample space, as shown in Fig. 1. Each event of interest corresponds to a leaf of the tree and its probability is equal to the product of the probabilities associated with the branches in a path from the root to the corresponding leaf. The desired probabilities of false alarm and missed detection are

P(false alarm)=P(Ac∩B)=P(Ac)P(B|Ac)=0.95∙0.10=0.095,
P(missed detection)=P(A∩Bc)=P(A)P(Bc|A)=0.05∙0.01=0.0005.

Application of Bayes` rule in this problem.
We are given that P(A)=0.05, P(B|A)=0.99, P(B|Ac)=0.1.
Applying Bayes’ rule, with A1=A and A2=Ac, we obtain

P(aircraft present | radar registers) = P(A|B)=PAPB APB=P(A)P(B |A))PAPB A+ PAcPB Ac==0.05∙0.990.05∙0.99+0.95∙0.1≈0.3426.

Problem 9 (Independence). Communication through a noisy channel. A binary (0 or 1) symbol transmitted through a noisy communication channel is received incorrectly with probability ε0 and ε1, respectively (see Fig. 2). Errors in different symbol transmissions are independent.

Figure 2: Error probabilities in a binary communication channel.

(a) Suppose that the channel source transmits a 0 with probability p and transmits a 1 with probability 1-p. What is the probability that a randomly chosen symbol is received correctly?
(b) Suppose that the string of symbols 1011 is transmitted. What is the probability that all the symbols in the string are received correctly?
(c) In an effort to improve reliability, each symbol is transmitted three times and the received symbol is decoded by majority rule. In other words, a 0 (or 1) is transmitted as 000 (or 111, respectively), and it is decoded at the receiver as a 0 (or 1) if and only if the received three-symbol string contains at least two 0s (or 1s, respectively). What is the probability that a transmitted 0 is correctly decoded?
(d) Suppose that the channel source transmits a 0 with probability p and transmits a 1 with probability 1-p, and that the scheme of part (c) is used. What is the probability that a 0 was transmitted given that the received string is 101?
Solution:
(a) Let A be the event that a 0 is transmitted. Using the total probability theorem, the desired probability is

PA1-ε0+1-P(A)(1-ε1)=p(1-ε0)+(1-p)(1-ε1).

(b) By independence, the probability that the string 1011 is received correctly is

1-ε0(1-ε1)3 .

(c) In order for a 0 to be decoded correctly, the received string must be 000, 001, 010, or 100. Given that the string transmitted was 000, the probability of receiving 000 is (1-ε0)3, and the probability of each of the strings 001, 010, and 100 is ε0(1-ε0)2,. Thus, the probability of correct decoding is

3ε0(1-ε0)2+(1-ε0)3.

(d) Using Bayes’ rule, we have

P(0 | 101)=P(0)P(101 | 0)P(0)P(101 | 0)+P(1)P(101 | 1).

The probabilities needed in the above formula are

P(0)=p, P(1)=1-p, P(101 | 0)=ε02(1-ε0), P(101 | 1)=ε1(1-ε1)2.

Problem 10 (RELIABILITY). An electrical system consists of identical components that are operational with probability p, independently of other components. The components are connected in three subsystems, as shown in Fig. 3. The system is operational if there is a path that starts at point A, ends at point B, and consists of operational components. This is the same as requiring that all three subsystems are operational. What are the probabilities that the three subsystems, as well as the entire system, are operational?

Figure 3: A system of identical components that consists of the three subsystems 1, 2, and 3. The system is operational if there is a path that starts at point A, ends at point B, and consists of operational components.

Solution:
The system may be viewed as a series connection of three subsystems, denoted 1, 2, and 3, in Fig. 3 in the text. The probability that the entire system is operational is p1p2p3, where pi is the probability that subsystem i is operational. Using the formulas for the probability of success of a series or a parallel system, we have

p1=p, p3=1-(1-p)2,

and

p2=1-(1-p)1-p1-(1-p)3.

Problem 11. Reliability of a k-out-of-n system. A system consists of n identical components that are operational with probability p, independently of other components. The system is operational if at least k out of the n components are operational. What is the probability that the system is operational?
Solution:
Let Ai be the event that exactly i components are operational. The probability that the system is operational is the probability of the union i=knAi, and since the Ai are disjoint, it is equal to

i=knP(Ai)=i=knp(i),

where p(i) are the binomial probabilities. Thus, the probability of an operational system is

i=knnipi(1-p)n-i.

Problem 12. A cellular phone system services a population of n1 “voice users” (those that occasionally need a voice connection) and n2 “data users” (those that occasionally need a data connection). We estimate that at a given time, each user will need to be connected to the system with probability p1 (for voice users) or p2 (for data users), independently of other users. The data rate for a voice user is r1 bits/sec and for a data user is r2 bits/sec. The cellular system has a total capacity of c bits/sec. What is the probability that more users want to use the system than the system can accommodate?
Solution:
The probability that k1 voice users and k2 data users simultaneously need to be connected is p1(k1)p2(k2), where p1(k1) and p2(k2) are the corresponding binomial probabilities, given by

pi(ki)=nikipiki1-pini-ki, i=1, 2.

The probability that more users want to use the system than the system can accommodate is the sum of all products p1(k1)p2(k2) as k1 and k2 range over all possible values whose total bit rate requirement k1r1+k2r2 exceeds the capacity c of the system. Thus, the desired probability is

{(k1,k2) | k1r1+k2r2>c, k1≤n1, k2≤n2}p1(k1)p2(k2).

Problem 13. Grade of service. An internet service provider has installed c modems to serve the needs of a population of n customers. It is estimated that at a given time, each customer will need a connection with probability p, independently of the others. What is the probability that there are more customers needing a connection than there are modems?
Solution:
Here we are interested in the probability that more than c customers simultaneously need a connection. It is equal to

k=c+1np(k),

where

p(k)=nkpk(1-p)n-k

are the binomial probabilities. For instance, if n=100, p=0.1, and c=15, the desired probability turns out to be 0.0399.
This example is typical of problems of sizing the capacity of a facility to serve the needs of a homogeneous population, consisting of independently acting customers. The problem is to select the size c to achieve a certain threshold probability (sometimes called grade of service) that no user is left unserved.