ECE 450 - csun.edudvanalp/ECE 450/ece_450_hw_perm/ece_450_hwset_1/ece_hw… · 8/24/2015 ECE 450 8...

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ECE 450

Homework Set 1 – Part 2 Solutions

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1. Pr(7 or 11) on toss of 2 dice

+ 1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

Pr(7 or 11) = 8/36 = 2/9

2. Pr(Q of ) in 5-card poker

Pr(Q of ) = 52/5

5

52

4

511

Numerator: 1 way to draw the

Q of ; 51C4 ways to draw the

“other” 4 cards in the hand

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3. 4 women check coats; randomly returned;

Pr(all 4 get their coats back)

Numerator: # of ways to correctly return coats = 1

Denominator: # of ways to return coats at random:

4! = 4 3 2 1 = 24

Pr(all 4 get their coats back) = 1/24

4. Passwords (single letter or letter followed by 3 symbols that are letters or digits)

Denominator: # of possible password:

26 + 26*36 + 26*36*36 + 26*36*36*36 =1,247,714

Pr(J & M same password) = 1/1,247,714 8.015*10-7

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5. Machine works if at least 6 of 7 parts work;

Pr(part failure) = .2; find Pr(machine failure)

Pr(failure) = 1 – Pr(at least 6 parts work)

= 1 – Pr(exactly 6 parts work or exactly 7 parts work)

= 1 – {Pr(exactly 6 parts work) + Pr(exactly 7 parts work)}

= 4233.)2(.)8(.7

7)2(.)8(.

6

71 0716

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6. Given: P(G) = P(B) = ½;

find Pr(G added| B sel.)

Bayes’ Rule: Pr(G added| B sel.)

= Pr(B sel.| G added) Pr(G added)/Pr(B sel.)

Total Prob. on Denominator, Pr(B sel.):

Pr(B sel.) = Pr(B sel.| Gadded) Pr(G added) + Pr(B sel.| B added) Pr(B added)

So,

Pr(G added| B sel.) = 2/(3+x)(1/2) / [ 2/(3+x)(1/2) + 3/(3+x) (1/2)] = 2/5

Now If G added If B added

# boys = #B 2 2 3

# girls = #G x x+1 x

Total # Babies 2+x 3+x 3+x

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8. Missile launches if both relays fail; Pr(A fails)

= .01, Pr(B fails) = .03, Pr(B fails|A fails) = .06;

a. Pr(launch) = Pr(A fails B fails)

= Pr(A fails) Pr(B fails|A fails) = .01(.06) = .0006

b. Pr(A fails|B fails) = (defn. cond. prob.)

= .0006/.03 = .02

c. Not independent since Pr(A fails|B fails) Pr(A fails)

)failsBPr(

)failsBfailsAPr(

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9. Pr(win) = .2 on each toss;

person A goes 1st

Let A denote the event that Person A wins on any particular toss

Let B denote the event that Person B wins on any particular toss

Possible ways for A to win the game:

– A (A wins on 1st toss)

• Probability: (.2)

– A’B’A (A wins on 3rd toss)

• Probability: (.8) (.8) (.2) = (.64) (.2)

– A’B’A’B’A (A wins on 5th toss)

• Probability: (.8) (.8) (.8) (.8) (.2) = (64)2 (.2)

– etc.

• Overall probability that A wins:

– Pr(A wins) = (.2) + (.64)(.2) + (.64)2 (.2) + …

= .2/(1-.64) = 5/9 .5556 Infinite geometric

series, r = .64

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10. Pr(spacecraft in zone) = .8;

6 spacecraft

a. Pr(none of 6 in zone) = Pr(0 successes, 6 failures)

b. Pr(at least 1 in zone) = 1 – Pr(all 6 out of zone)

= 1 - .000064 = .999936

c. Pr(3 or more in zone) = 1 – Pr(all out) – Pr(5 out, 1 in)

– Pr(4 out, 2 in)

= .999936 - ___________ - __________ = _____

000064.)2(.)8(.0

6 60

.999936

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11. (Haddad) Suppose: 30% of the packets are routed via Channel A;

1% of the packets are lost on Channel A; and

0.5% of the packets are lost on Channel B.

a. If one packet is sent, find the probability that it is lost.

b. If a packet is lost, what is the prob. that it was sent via Channel A?

c. If three packets are sent, what is the probability that all three packets

are lost?

a. Pr(loss) = Pr(loss|A)Pr(A) + Pr(loss|B)Pr(B) (Total Prob. Eq.)

= .01 (.3) + .005 (.7) = .003 + .0035 = .0065

b. Pr(A|loss) = .4615

c. Pr(3 losses in 3 trials) =

= 2.746 x 10^-7

0065./003.0065.

)3(.01.

)lossPr(

)APr()A|lossPr(

303 )0065(.)9935(.)0065(.3

3

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12. 26 letters written at random; Pr(x, y adjacent)

___ ___ ___ . . . ___ ___ ___

1 2 3 24 25 26

But it can be “x y” or “y x”, so there are 50 ways to place the x & y

adjacently, with 24! ways to fill in the other letters:

Pr(x, y adjacent) =

# of adjacent pairs of slots

to place the x & y: 25

13/12526

50

!26

!2450

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13. Pr(flush) in 5-card poker hand

Numerator = # of ways to get a flush

= (# of ways to get 5 spades) + (# of ways to get 5 clubs) +

(# of ways to get 5 hearts) + (# of ways to get 5 diamonds)

=

Denominator = # of possible 5-card poker hands =

Pr(flush) = .00196

5

134

5

13

5

13

5

13

5

13

5

52

5

52

5

134

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14. Eng. spelling: “rigour”; Am. spelling: “rigor”;

select letter at random; Pr(Eng) = .4, Pr(Am) = .6

Bayes: Pr(Eng | vowel) =

=

)vowelPr(

)EngPr()Eng|vowelPr(

...4545.11/5)5/3)(5/2()5/2()2/1(

)5/2()2/1(

Total

Probability on

Denominator

)AmPr()Am|vowelPr()EngPr()Eng|vowelPr(

)EngPr()Eng|vowelPr(

17. K = “knows”, K’ = “doesn’t know”;

R = “gets it right”:

Givens: Pr(K) = ¾; Pr(K’) = ¼

Pr(R|K’) = ¼ Pr(R|K) = _____

Find: Pr(K|R) = _____________ / ______ (Bayes’)

= _____________/ [ _______ ____ + ________ ___]

= (1)(3/4) / [(1)(3/4) + (1/4)(1/4)]

= (3/4) / [3/4 + 1/16] = 12/(12+1) = 12/13

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18. Box contains 15 sections, 5 defective;

choose 3; Pr(successful communication) = ?

Numerator = # of ways to choose 3 good links from 10

Denominator = # of ways to choose 3 links from all 15 in box

Pr(successful comm.) =

19. Family of 5 childrena. Pr(all same sex) = Pr(all boys) + Pr(all girls)

= (1/2)5 + (1/2)5 = 1/16

2637.91/24)123/(131415

)123/(8910

3

15

3

10

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19. Family of 5 children

b. Pr(B B B G G) = (1/2)5 = 1/32

c. Pr(exactly 3 boys) =

d. Pr(at least 1 girl) = 1 – Pr(all 5 are boys)

= 1 – (1/2)5 = 31/32

16/532

10)2/1()2/1(

3

5 23

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20. Pr(node failure) = .2

a. Pr(comm) = Pr{A D (B C)}

= Pr(A) Pr(D) Pr(B C) since events are independent

= (.8) (.8) {Pr(B) + Pr(C) – Pr(B C)}

= .64 (.8 + .8 – (.64)) = .6144

b. Pr(B’|comm) =

A

B

D

C

Computer 1 Computer 2

1666.6144.

)2(.8.

6144.

)2(.)DCAPr( 3

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22. Verify Part II, Prob. 1: Pr(7 or 11)

MATLAB Code for 10,000 Trials

% program ece_450_hw1

red = randint(10000, 1, [1 6]);

green = randint(10000, 1, [1 6]);

dice = red + green;

flag = (dice == 7) | (dice == 11);

count = 0;

for i = 1:10000

count = count + flag(i);

end;

prob = count/10000

# Trials Pr(7 or 11)

10 .2

100 .25

1000 .2

10,000 .226

Theoretical

Ans.

2/9 = .2222

Note: You may have answers

that vary somewhat from these;

however, for 10,000 trials your

answer should be close to the

theoretical answer.

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23. Roll a die; N = # showing; toss coin N

times; Y = # heads showing; find Pr(Y = 0),

Pr(Y = 3); Matlab Simulation Code:

% program ece_450_hw1_prob23

dice = randi([1 6 ] [10000,1]);

Y = zeros(10000,1); % initialize Y

for i = 1:10000 % 10K trials

Y(i) = 0;

N = dice(i);

coin = randi([0 1], [N,1] )% 1 = heads

for j = 1:N

Y(i) = Y(i) + coin(j);

end

end

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23. Roll a die; N = # showing; toss coin N

times; Y = # heads showing; find Pr(Y = 0),

Pr(Y = 3)

MATLAB Code, continued

flag0 = (Y == 0); flag3 = (Y == 3);

count0 = sum(flag0);

count3 = sum(flag3);

prob0 = count0/10000

prob3 = count3/10000

Command Window:

>> ece_450_hw1_prob23

prob0 =

0.1655

prob3 =

0.1680

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24. Try adding in a 3rd parallel path, through

some node E

Pr(comm) = Pr{A D [(B C) E]}

= Pr(A) Pr(D) Pr[(B C) E] by independence

= (.8) (.8) {Pr(B C) + Pr(E) – Pr[(B C) E]}

= .64 {.96 + .8 - .96(.8)} = .6349 > .62

redundancy 3 is sufficient

A

B

D

C

Computer 1 Computer 2

E

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25.

Pr(Error) = Pr(E) = Pr(E | 0t) Pr(0t) + Pr(E | 1t) Pr(1t)

= (1-p) (.5) + (1-p) (.5) = 1 – p .001

p .999

.5

.5

p

p

1-p

1-pFind minimum

acceptable p to get P(E)

.001

(Total Probability Formula)

26. (Cooper & McGillem 1-4.8) Givens: Pr(truck) = ¼;

Pr(unsafe|auto) = 1/8; Pr(unsafe|truck) = 1/20.

a) Pr(truck unsafe) = Pr(unsafe|truck) Pr(truck)

= (1/20) (1/4) = 1/80

b) Pr(truck|unsafe) = Pr(unsafe|truck) Pr(truck)/P(unsafe)

by Bayes’ Rule

Total Prob. on Denom:

Pr(unsafe) = Pr(unsafe|truck) Pr(truck)

+ Pr(unsafe|auto) Pr(auto)

= (1/20) (1/4) + (1/8) (3/4) = 1/80 + 3/32

= 2/160 + 15/160 = 17/160

So, Pr(truck|unsafe) = 2/17

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26 continued

c) Pr(truck|previous auto) = Pr(truck) = ¼

(assumes that the event {truck} and the event {car} are

independent)

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27. Cooper & McG 1-6.3: Randomly draw 3 cards in

succession from deck of 52; A = King on 1st draw, B = King

on 2nd draw, C = King on 3rd draw

a) A B’ : the event of drawing a King on the 1st draw and

drawing anything but a King on the 2nd draw

Pr(A B’) = Pr(A) Pr(B’|A) = (1/13) (48/51) = .0724

b. A B : the event of drawing a King on the 1st or 2nd draw;

Pr(A B ) = Pr(A) + Pr(B) – Pr(A B)

= (1/13) + (1/13) – Pr(A) Pr(B|A)

= (2/13) – (1/13)(3/51) = .1493

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d. A’ B’ C’ : the event of drawing none-King on the 1st,

2nd, and 3rd draw

Pr( A’ B’ C’ ) = Pr( (A’ B’) C’)

= Pr((A’ B’)) Pr(C’| (A’ B’) ) **

where Pr(A’ B’) = Pr(A’) Pr(B’|A’)

= (48/52) (47|51) = (12/13)(47/51)

and Pr(C’| (A’ B’) ) = 46/50 = 23/25

Plugging into **: Pr( A’ B’ C’ ) = (12/13)(47/51)(23/25)

= .7826

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Remaining answers:

c. .99548

e. .07692

f. .93195

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28. Cooper & McG, 1-6.5 Two diodes connected in series;

Each has Pr(fails as short) = .05, Pr(fails as open) = .1;

Assume independence; find Pr(series connection functions

as diode).

• Pr(functions as diode) = Pr(both work or (one works and

the other fails as a short)), m.e.

= Pr(both work) + Pr(one works and the other fails

as a short)

= .852 + (.85) .05 + .05 (.85) = .8075

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33. Cooper & McGillem, problem 1-10.8a: 10,000

characters are tx’d as a file; Pr(error) = .001 per character;

find Pr(file is tx’d without error).

Pr(file tx’d without error) = Pr(all characters tx’d correctly)

= (.999)10000 = 4.517 x 10 ^(-5)

36. (Lecture 1; Mix) Toss a die, so the sample space is the

set of numbers 1 – 6. Find the entropy in this experiment.

H = average info/event

= 2.585

Events Pr(event) I(event)

1 1/6 Log2(6) = 2.5850

…

6 1/6 Log2(6) = 2.5850

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35. (Lecture 1)

a. Consider a quaternary alphabet with 4 equally-likely symbols: {0, 1, 2, 3).

Find the entropy of the alphabet.

b. Find the entropy of the same alphabet if the symbol probabilities are:

Pr(0) = Pr(1) = 1/4; Pr(2) = 1/8; Pr(3) = 3/8. Also determine which of the 4

symbols carries the most information.

a. Information in each of the four symbols: log2(4) = 2 bits; thus H =

average information per symbol = 2.bits

b.

H = ¼ (2 bits) + ¼ (2 bits) + 1/8 (3 bits) + 3/8 (1.415 bits) = 1.9056 bits

Symbol Pr(symbol) I(symbol)

0 ¼ Log2(4) = 2 bits

1 ¼ Log2(4) = 2 bits

2 1/8 Log2(8) = 3 bits

3 3/8 Log2(8/3) = 1.415 bits

2 carries the most

information, since it is

the least likely.

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