EEL 3470: Electromagnetic Fields and Waves

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Dr. P. S. NeelakantaEEL 4541: STOCHASTIC PROCESSES & RANDOM SIGNALSUNIT A-II Self-study Homework Assignments with Solutions

________________________________________________________________________

Topics: RANDOM SAMPLE SPACE: Deterministic and random variables (RVs)

OCCURRENCE OF RANDOM EVENTS

STATISTICAL EXPERIMENTS: Counting the chances of occurrence

PROBABILITY CONCEPTS Definition; types; rules

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RANDOM AND DETERMINISTIC EVENT SPACES

Events may occur randomly. They can be represented by random variables (RVs). The likelihood of occurrence of such random events can be evaluated via statistical experiments (trials).

The likelihood of occurrence can be quantified by a set of real numbers (called weights on the likelihood) denoting the probability of occurrence (P) that ranges from 0 to 1.

Probability of an event A: 0 ( P(A) ( 1; P(() = 0; P(S) = 1

Axioms of probability: (i) 0 ( P(Xi) ( 1; (ii)

P(D)0.01

=

; (iii)

i

P(X)

: Never less than zero; and, (iv) Xi should be exhaustive namely, inclusive of all events i = 1, 2, N that exist in the event space

Example

When a coin is tossed twice, the events that may happen are: HH, HT, TT, TH with each pair having a likelihood of occurrence .

Relevant to this event space of S = {HH, HT, TT, TH}, suppose an exhaustive experiment of coin-tossing is performed and pairs of events are observed. If A denotes the event of at least one H occurring, then, A {HH, HT, TH} and P(A) = (3/4) = ( + + = .)

Equiprobable epochs

If an experiment can result in any one of N different equally likely outcomes, and if exactly n of these outcomes correspond to event A, then the probability of event A is: P(A) = n/N.

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Problem A-II-1

A mixture of resistors in a box contains items of six -watt, four -watt and three 1-watt ratings. If a random selection is made blindly from this mixture, find the probability of: (i) getting a (1/4-watt) rated entity; and (ii) getting either a (1/2-watt) or a (1-watt) rated entity.

Solution:

Total items: 13

(i) Selecting -watt item in random: P(-watt resistor) = 6/13

(ii)Since seven of 13 items are either 1/2-watt or 1-watt resistors),

P(1/2-watt resistor ( 1-watt resistor) = 7/13

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PROBABILITY RULES/TYPES

(a) Additive rule: Applies to union of events

If A and B are any two events, then P(A(B) = P(A) + P(B) P(A(B)

i

i

P(X)=1

If A and B are mutually exclusive: P(A(B) = P(A) + P(B)

(That is, when no intersection of A and B prevails)

P(D)0.01

=

P(D)0.99

=

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Problem A-II-2

For a system, the probability that it passes the DC bias conditions is: 2/3; and, the probability that it passes environmental testing is 4/9. Suppose the probability that the system passes both is , what is the probability that the system survives passing at least one of these tests?

Solution

Event passing DC conditions: dcT

Event environmental testing: enT

P(dcT ( enT) = P(dcT) + P(enT) P(dcT ( enT) =(2/3) + (4/9) (1/4) = 31/36

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Problem A-II-3

At a traffic-signal, the probabilities 0.09, 0.15, 0.21 and 0.23 denote respectively, the individual statistics of non-operational (N), amber (A), red (R) and green (G) states of the signal. Find the probability of observing any one of the states as above.

Solution:

All events are mutually exclusive.

So, P(N (A(R(G) = (0.09 + 0.15 + 0.21 + 0.23) = 0.68

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(b) On complementary events P(A) + P(A) = 1

Problem A-II-4

If power outage occurs at 3, 4, 5, 6, 7 and (8 or more) locations respectively with probabilities: 0.12, 0.19, 0.28, 0.24, 0.10 and 0.07 respectively, what is the probability that power outage at least in 5 locations can be expected?

Solution:

Let O5 denotes the event that at least 5 outages happen. Now, P(O5) = 1(P(O5) where O5 is the event that fewer than 5 outages take place. So, P(O5) = 0.12 + 0.19 = 0.31. Therefore, P(O5) = 1 ( P(O5) = (1 0.31) = 0.69.

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(c) Self (a priori) probability, conditional (a posteriori) probability and joint probability

Self (a priori) probability: This refers to the likelihood of occurrence of an event A regardless of any other events posing influence of this event. That is, A occurs all by itself with a self probability P(A) unconditioned.

On the other hand, suppose an event B occurs subject to the condition that A has occurred. That is, the probability P(B|A) depicts a conditional (a posteriori) probability namely, the probability that B occurs given that A occurs.

Joint probability: Joint probability of say, 2 events X and Y refers to the chance of the epochs X and Y occurring together. It can be specified via Bayes rule:

P(X, Y) = P(X)P(Y|X) = P(Y)P(X|Y)

Problem A-II-5

Suppose a radar receiver input (Y) denotes either signal-plus-noise (target-plus-clutter) or just noise (N) denoting clutter alone. Determine the posterior probability: P[(X = S+N)|Receiver input: Y].

Solution:

The probability of the event Y (that is occurrence of an input at the receiver) is:

P(Y) = P(Y|X)P(X) + P(Y|N)P(N)

Therefore,

P[(X = S+N)|Receiver input: Y]= [P(X) P(Y|X)]/ [P(Y|X)P(X) + P(Y|N)P(N)]

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Problem A-II-6

A coin with head (H) on one side and tail (T) on the other side is flipped repeatedly to a statistically large extent of times. In this random experiment, which of the following refers to the correct set of possible outcomes anticipated and the probabilities of successive occurrences of H and T: (a) {TH, HT: 0.5, 0.5}; (b) {TH, HT: 0.4, 0.6}; (c) {TH, HT, TT: 1/3, 1/3, 1/3}; (d) {TH, HT, TT, HH: , , , }; and (e) {TH, HT, TT, HH: , , , }. Why?

Solution: Answer (e): {TH, HT, TT, HH: , , , }. In statistically large possible events, H and T would pair up successively as {TH, HT, TT, HH}with the four elements denoting all possible combinations exhaustively. Further, they are equally-likely events.

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Problem A-II-7

If four events E

1

, E

2

,

E

3

and E

4

are mutually exclusive of a random experiment, which of the following denotes proper assignments of possibility values? Why?

(a) P

(

)

1

E

= 1.5; P

(

)

2

E

= ( 0.5; P

(

)

3

E

= P

(

)

4

E

= 0

(b) P

(

)

1

E

EMBED Equation.3 = 0.5; P

(

)

2

E

= 0.25; P

(

)

3

E

= 0.25; P

(

)

4

E

= 0.25

(c) P

(

)

1

E

= 0.5; P

(

)

2

E

= 1/3; P

(

)

3

E

= 1/6; P

(

)

4

E

= 0

(d) P

(

)

1

E

=

2

1

; P

(

)

2

E

= -

5

1

EMBED Equation.3 ; P

(

)

3

E

= P

(

)

4

E

=

2

1

(e) P

(

)

1

E

= ( 0.1; P

(

)

2

E

= + 0.1; P

(

)

3

E

= 0.5; P

(

)

4

E

= 0.4

Solution: Answer (c): p(E1) = ; p(E2) = 1/3; p(E3) = 1/6; p(E4) = 0.

Only (c) satisfies all the axioms of probabilities, namely:

(i) 0 p(Ei) 1,i = 1, 2, 3, 4

(ii) p(Ei) = 1

(iii) p(Ei) should never be less than 0

(iv) The probability count is exhaustive across the sample space

[(a), (b), (d) and (e) fail to satisfy at least one of the axioms]

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Problem A-II-8

A robot is designed to select an item randomly from a set of 10 units serial numbered as: 00, 01,, 09. Find the probability of the following event: (i) if each event is equally likely to occur and the serial number of the unit selected is less than 5. Choose the correct answer: (a) 5/9; (b) 0.25; (c) 0.33; (d) 0.5; (e) 1.0. Justify the choice of your answer.

Solution: Answer (d): 0.5

There are 10 numbers (0-9). Suppose inn selecting (randomly) an item with a serial number less than 5 is designated as C. This sample space C is :{0, 1, 2, 3, 4} and it has 5 elements. Therefore, p(C) = 5/10 = 0.5.

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Problem A-II-9

In Q # Problem A-II-8, find the probability of the following events, if each event is equally likely to occur.

(i) Serial number of the unit selected is odd.

(ii) Serial number of the unit selected is a multiple of 3.

(iii) Serial number of the unit selected is even.

(iv)Serial number of the unit selected is either odd or a multiple of 3.

(v) Serial number of the unit selected is either odd, or multiple of 3, or is less than 5.

Solution:

(i) Serial number of the unit selected is odd ( {A} = {1, 3, 5, 7, 9}. (P(A) = 5/10

(ii) Serial number of the unit selected is a multiple of 3. ( {B} = {3, 6, 9}. (P(B) = 3/10

(iii) Serial number of the unit selected is even. ( {D} = {0, 2, 4, 6, 8}. (P(D) = 5/10

(iv) Serial number of the unit selected is either odd or a multiple of 3. ( P{A ( B} = P{A}+ P{B}( P{A ( B}= 5/10 +3/10 ( 2/10 = 6/10.

(v) Serial number of the unit selected is either odd, or multiple of 3, or is less than 5: P(A U B U C) = P(A) + P(B) + P(C) P(AB) P(BC) P(AC) + P (ABC) = 5/10 + 3/10 + 5/10 2/10 2/10 1/10 - 1/10 = 9/10

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Problem A-II-10

Two components in electronic circuit C

1

and C

2

are tested and declared to be in one of the following 3 possible states:

State F: Both C

1

and C

2

are functioning

State R: Both are not functioning but repairable

State K: Both are damaged beyond repair

What is the set corresponding to the event none of the components (C1 or C2) in state of K? Choose the correct answer: (a) {FF, RR}; (b) {FR, RF}; (c) {FF, RR, FR, RF}; (d) {FF, RR, FR}; (e) {FF, RR, RF}. Why?

Solution: Answer (c): {FF, RR, FR, RF}

For the event, none of the component (C1 or C2) is in the state K, the sample space can be specified with the omission of K. That is:

P(F/D)0.02

=

C1

C2

F

R

F

FF

FR

R

RF

RR

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Problem A-II-11

Reference to Q # Problem A-II-10, what is the sample space in the experiment, considering exhaustively all possible states of C

1

and C

2

?

Solution: The entire sample space (that includes all the events F, R and K) is as follows:

P(F/D)0.05

=

C2

C1

F

R

K

F

FF

FR

FK

R

RF

RR

RK

K

KF

KR

KK

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Problem A-II-12

For the asymmetric binary channel shown, assuming that the occurrence of input events (0, 1) are equiprobable.

(i) Find the probability of occurrence of zeros at the output.