Solution Manual for Introduction to Probability Models 7th

Solution Manual for:

Introduction to Probability Models : 7th

by Sheldon M. Ross.

John L. Weatherwax∗

January, 2012

By Mathematics 5F

UHAMKA

Solution Manual for : Introduction to Probability Models

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1. An urn contains five red, three orange and two blue balls. Two balls are randomly

selected. What is the sample space of this experiment ? Let X represent the number of

orange balls selected. Where are the possible values of X ? Calculate P {X = 0}.

Solutions :

The sample space of this experiment will contain elements that are 45 elements. It’s

will be like

{

[(

) ] [ ]

[ ] [( ) ]

[ ] [( ) ]

}

If X represents the number of orange balls selected

Than X can have three values

That is X = 0, 1, 2

2. Let X represent the difference between the number of heads and the number of tails

obtained when a coin is tossed n times. What are the possible values of X ?

Solutions :

3. In exercise 2, if the coin is assumed fair, then for n = 2, what are the probabilities

associated with the values that X can take on ?

Solutions :

for n= 2 sample space will be S = {HH, HT, TH, TT}


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‘X’ can take on value 0 or 2 P(X=0) = P{HT, TH}

P(X=2) = P = {HH,TT} =

4. Suppose a die is rolled twice. What are the possible values that the following random

variables can take on ?

(i) The maximum value to appear in the two rolls.

(ii) The minimum value to appear in the two rolls.

(iii) The sum of the two rolls.

(iv) The value of the first roll minus the value of the second roll.

solution:

(i) 1, 2, 3, 4, 5, 6

(ii) 1, 2, 3, 4, 5, 6

(iii) 2, 3, 4, 5, 6, 7, 8. 9, 10, 11, 12

(iv) -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5

5. If the die in Exercise 4 is assumed fair, calculate the probabilities associated with the

random variables in (i) – (iv).

solution:

Step 1

(i)

p (x = 1) = p {(1,1)} =

p (x = 2) = p {(2,1) (2,2) (1,2)} =

p (x = 3) = p {(3,1) (3,2) (3,3) (2,3) (1,3)} =

p (x = 4) = p {(4,1) (4,2) (4,3) (4,4) (3,4) (2,4) (1,4)} =

p (x = 5) = p {(5,1) (5,2) (5,3) (5,4) (5,5) (4,5) (3,5) (2,5) (1,5)} =

p (x = 6) = p {(6,1) (6,2) (6,3) (6,4) (6,5) (6,6) (5,6) (4,6) (3,6) (2,6) (1,6)} =


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Step 2

x 1 2 3 4 5 6

p (x)

Step 3

(ii)

p (x = 1) = p {(6,1) (5,1) (4,1) (3,1) (2,1) (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)} =

p (x = 2) = p {(6,2) (5,2) (4,2) (3,2) (2,2) (2,3) (2,4) (2,5) (2,6)} =

p (x = 3) = p {(6,3) (5,3) (4,3) (3,3) (3,4) (3,5) (3,6)} =

p (x = 4) = p {( (6,4) (5,4) (4,4) (4,5) (4,6) )} =

p (x = 5) = p {( (6,5) (5,5) (5,6) )} =

p (x = 6) = p {(6,6)} =

Step 4

x 1 2 3 4 5 6

p (x)

Step 5

(iii)

p (x = 2) = p {(1,1)} =

p (x = 3) = p {(1,2) (2,1)} =

p (x = 4) = p {(1,3) (2,2) (3,1)} =

p (x = 5) = p {(1,4) (2,3) (3,2) (3,1)} =

p (x = 6) = p {(1,5) (2,4) (3,3) (4,2) (5,1)} =

p (x = 7) = p {(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)} =


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p (x = 8) = p {(2,6) (3,5) (4,4) (5,3) (6,2)} =

p (x = 9) = p {(3,6) (4,5) (5,4) (6,3)} =

p (x = 10) = p {(4,6) (5,5) (6,4)} =

p (x = 11) = p {(5,6) (6,5)} =

p (x = 12) = p {(6,6)} =

Step 6

x 2 3 4 5 6 7 8 9 10 11 12

p (x)

Step 7

(iv)

p (x = 5) = p {(6,1)} =

p (x = 4) = p {(6,2) (5,11)} =

=

p (x = 3) = p {(6,3) (5,2) (4,1)} =

=

p (x = 2) = p {(6,4) (5,3) (4,2) (3,1)} =

=

p (x = 1) = p {(6,5) (5,4) (4,3) (3,2) (2,1)} =

p (x = 0) = p {(6,6) (5,5) (4,4) (3,3) (2,2) (1,1)} =

=

p (x = -1) = p{(5,6) (4,5) (3,4) (2,3) (1,2)} =

p (x = -2) = p {(4,6) (3,5) (2,4) (1,3)} =

=

p (x = -3) = p {(3,6) (2,5) (1,4)} =

=

p (x = -4) = p {(2,6) (1,5)} =

=

p (x = -5) = p {(1,6)} =


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6. Suppose five fair coins are tossed. Let E be the event that all coins land heads. Define

the random variable EI

1, if E occurs

EI =

0, if cE occurs

For what outcomes in the original sample space does EI equal 1 ? What is 1EP I ?

solution :

Ruang Sampel :

H= HHHHH =

H

T = HHHHT

H

H = HHHTH

T

T = HHHTT

H

H = HHTHH

H

T = HHTHT

T

H = HHTTH

T

T = HHTTT

H

H = HTHHH

H

T = HTHHT

H

H = HTHTH

T

T = HTHTT

T

H = HTTHH

H

T = HTTHT

T

H = HTTTH

T

T = HTTTT


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H= THHHH

H

T = THHHT

H

H = THHTH

T

T = THHTT

H

H = THTHH

H

T = THTHT

T

H = THTTH

T

T = THTTT

T

H = TTHHH

H

T = TTHHT

H

H = TTHTH

T

T = TTHTT

T

H = TTTHH

H

T = TTTHT

T

H = TTTTH

T

T = TTTTT

1 , = 132

Jadi , EI dan 1 0P P

0 , = 31

32

1 31 321

32 32 32


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7. Suppose a coin having probability 0,7 of coming up heads is tossed three times. Let X

denotethe number of heads that appear in the three tosses. Determine the probability

mass function of X.

solution:

3

2

2

3

(0) (0,3) 0,027

(1) 3(0,3) (0,7) 0,189

(2) 3(0,3)(0,7) 0,441

(3) (0,7) 0,343

(0) (1) (2) (3) 0,027 0,189 0,441 0,343 1

P

P

P

P

P P P P

8. SOAL ?

Solution : 11

00

1 1 1

2 2 2dx x

1

0

1 1dx x

Jadi 1 3

( ) 12 2

P X

9. If the distribution function of F is given by

0, 0

1, 0 1

2

3, 1 2

5

4, 2 3

5

9, 3 3,5

10

1, 3,5

b

b

b

F b

b

b

b

Calculate the probability mass function of X

Solution:


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1 10 0

2 2

3 1 11

5 2 10

4 3 12

5 5 5

9 4 13

10 5 10

9 13.5 1

10 10

P

P

P

P

P

10. SOAL ?

Solution :

P(paling banyak 1) = 1 1 1 1

. .6 6 6 36

P(paling banyak 6) = 6 6 6

. . 16 6 6

P(paling banyak 1 dan 6) = 1 1

.136 36


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11. A ball is drawn from an urn containing three white and three black balls. After the ball is

drawn, it is then replaced and another ball is drawn. This goes on indefinitely. What is the

probability that of the first four balls drawn, exactly two are white?

Solution

Peluang terambilnya 4 bola, dan selalu terdapat 2 bola putih

: 2

4

1

2

Diketahui i

n

p

1n ii

np i p p

i

2 4 2

2 2

4 1 12 1

2 2 2

4 1 1 3

2 2 2 8

p

12. on a multiple-choice exam with three possible answer for each of the five questions, what

is the probability that a student would get four or more correct answer just by guessing?

Diketahui: banyaknya pertanyaan = 5 dan peluang jawaban yang benar : ( )

( ) ( ) ( )

Keterangan:

: banyaknya pertanyaan

: banyaknya pertanyaan yang benar

: peluang jawaban yang benar

( ) : peluang jawaban yang benar

Solution:

Jika hanya mendapatkan 4 jawaban benar

( ) ( ) (

)

(

)


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( ) (

) (

)

( )

Jika mendapatkan 5 jawaban benar

( ) ( ) (

)

(

)

( ) (

)

( )

Jadi peluang seorang siswa mendapatkan 4 atau 5 jawaban yang benar adalah

( ) ( )

13. An individual claims to have extrasensory perception (ESP). As a test, a fair coin is

fliepped ten times, and he is asked to predict in advance the outcome. Our individual gets

seven out of ten correct. What is the probability he would have done at least this well if he

had no ESP? (Explain why the relevant probability is I2{X ≥ 7) and not P{ x = 7}.)

Solution

10

0

10

7

10 1

2

i

i

P i

i

14. SOAL ? 3 3

6 1 1( 3)

3 2 2P x

6! 1 1.

3!3! 8 8

6.5.4.3!

3!.3.2.1

120.

64

20 5

64 16


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15. Let be binomially distributed with parameter and . show tahat as goes from to

( ) increases monotonically, then decreases monotonically reaching its largest

value.

a. in the case that ( ) is an integer, when k equal either

( ) ( ) .

b. in the case that ( ) is not an integer, when k satisfies

( ) ( ) .

Solution

Let X be binomially distributed with parameter and , then

( )

( )

( )

(

)

( )

( ) ( )

( )

( )

( ) ( )

( )

( )

Mode is the value of for which ( ) is maximum. Then there are two cases arrises :

a. in case, ( ) is an integer

Let ( ) (an integer) ( )

( )

( )

( )


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from ( ) it obvious that

( )

( )

Hence if ( )is integer ( ) then ( ) inreases till and

( ) ( ) and after that it begins to decreases.

b. in case, ( ) is not an integer

Let ( ) +y (where is an integer

and is fractional) Such

Hence from (2)

( )

( )

( )

( )

from ( ) it is obvious that

( )

( )

Flence, if ( ) is not integer ( ), then ( ) increases till

and after it begins to decreases

16. An airline knows that 5 percent of the people making reservation on a certain fight will

not show up. Consequently, their policy is to sell 52 tiket for a flight that can only hold 50

passengers. What is the probability that there will be a seat available for every passenger

who shows up?

Solution

probability the people making reservation not shows up

probability the people making reservation shows up

the people making reservation


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passengers

∑ ( )

∑ ( )

∑ ( )

∑ ( )

∑ ( )

( ) (

)

∑ ( )

( )

( )

∑ ( )

( ) ( ) ( )

∑ ( )

the probability that there will be a seat available for every passenger who shows up is

17. suppose that an experiment can result in one of r possible outcomes, the ith outcomes

having probabilities pi, I =1, … , r, ∑ 1= 1. If n of these experiments are performed, and if

then show that the probability that the first outcome appears x1 times, the second x2 times,

and the rthxrtimes is

1 2

1 2

!

! !... !

x

r

np p

x x xwhen 1 2 ... rx x x n

This is known as the multinomial distribution

Solution:

Follows since since there are

permutations of n objects of which are a like, are a

like, ..., x, are a like.

18. show that when r= 2 the multinomial reduces to the polynomial

Answer: ?


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19. in exercise 17, let x1 denote the number of times the ith outcomes appears, i = 1, …, r.

what is the probability mass function of 1 2 ... kx x x ?

Solution:

{ } [

] ( )

( )

20. 50% purchase an ordinary television set.

20% purchase a color television set.

30% just be browsing.

Five cuctomers = one customers purchase an ordinary set + two customers purchase

color television set + two customers purchase nothing.

( ) ( ) ( )

For 50% >> ( ) ( ) (

)

(

)

(

) (

) (

)

For 20% >> ( ) ( ) (

)

(

)

(

) (

)

For 30% >> ( ) ( ) (

)

(

)

(

) (

)


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21. SOAL ?

Solution :

Let P denote the probability of persons purchasing any type of television sets then,

,

Let X denotes the no. of televisions sold. Where ‘X’ is a binomial random variable then

required probability is given by

{ } { } { }

( ) (

)

(

)

( ) (

)

(

) (

) (

)

(

)

[ ]

[ ]

[ ]

Hence, required probability

22. Soal ?

P(H pada lemparan kelima) =

51 1

2 32

23. A coin having probability p of coming up heads is successively flipped until the rth head

appears. Argue that X, the number of flips required, will be n, n ≥ r, with probability

P {X=n} = 1

1 ,1

n rrn

p p n rr

This is known as the negative binomial distribution.

Solutions :

The no of flips required to obtain the rth

succes a sequence succes in a sequence o

Independent Bernoulli trails then given probability is

P (X = n) = (

) pr (1- p)

n-r, n ≥ r

Is correct because to get rth

head in excatly nth

trial head should have appeared r – 1

times on n – 1 trials


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1 1( 1)

1 1 1

( ) . ( 1)

1( ) (1 )

1

1. (1 )

1

1( ) (1 )

1

n n r

r n r

r n r

P X n P P X n

nP X n P P p

r

nP P p

r

nP X n P p

r

So, it is proven.

24. the probability mass function of X is given by

P(k)= (

)pr(1-p)

n-r , n

Answer:

In Exercises 25 and 26, suppose that two teams are playing a series of games, each of which

is independently won by team A with probability p and by team B with probability 1 – p. The

winner of the series is the first team to win i games.

25. If i = 4, find the probability that a total of 7 games are played. Also show that this

probability is maximized when p =

ANSWER

The probability that the series and after 7th

game can be found if neither of the teams

have won after playing six games. It means that both have won 3 games each. Now, the

required probability is given by:

p = probability that A win at 7th

game or B wins at 7th

game.

= ( ) ( ) ( ) (

) ( )

= ( ) ( ) [ ]

OR

p = ( ) ( )


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= ( ) ( )

That, the required probability.

Now this probability depends on the value of p (1 – p). To find maximum value of

probability, we will have to find maximum value of p (1 – p).

Differentrate p (1 – p) with respect to p

= - p + 1 – p

= 1 – 2p

Put the above expression to zero

1 – 2p = 0

P =

Hence, the Exp. p (1 – p) is maximum at

.

Therefore, the required probability is maximum at

.

Hence, is proven.

26. Find the expected number of games that are played when

a) i = 2

b) i = 3

In both cases, show that this number is maximized when p =

ANSWER:

Let N be the number of games played.

a) i = 2

A wins 2 games out of 2 → p2

B wins 2 games out of 2 → (1 – p)2

A wins 2 games out of 3 → ( ) {( ) }

B wins 2 games out of 3 → ( ) {( )}

[ ]


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( ( ) )

( ( ) {(

) } ( ) {(

) })

To find the maximum number of games,

[ ] → p =

b) i = 3

A wins 3 games out of 3 → p3

B wins 3 games out of 3 → (1 – p)3

A wins 3 games out of 4 → ( ) {( ) }


A wins 3 games out of 5 → ( ) {( )}


[ ]

= ( ( ) ( ( ) {( ) } ( )

{( ) })

( ( ) {( ) } ( ) {(

) })

=

To find the maximum number of games,

[ ]

→

Then

, and we will refuse the other 2 values because one of them is negative and

the order is greater than 1.

27. A fair coin is independently flipped n times, k times by A and n-k times by B. Show that

the probability that A and B flip the same number of heads is equa to the probability that

there are a total of k heads.

Solutions :

Let


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Now, probability that A and B flip same no. Of heads is given by P {X=0, X=1, .... X=K}

Where X is a Random variables denothing numbers of heads therefor.

0 0 1 1 1

0 0 1 1

0 0 1

0 0 1

1 1 1 1 1 1 1 1 1 1. . ... .

2 2 2 2 2 2 2 2 2 2

1 1 1 1 1 1.

2 2 2 2 2 2

k n k k n k k n k

k n k k n k k n k

K k

k n k

k n k k

P C C C C C C

P C C C

1 1

1

1 1.

2 2

k n k

n kC

Where we know that probability that a flip of coin result in head is

and

( )

0 0 1 1

1. ...

2

n

k n k k n k k n k

k kP C C C C C C

Writing the above sequence in reverse order.

0 1 1 1 1 0 0

1. ...

2

n

k n k k n k k n k k n k

k k kP C C C C C C C C

Also we know that, .n n

r n r

Therefore, the above equation becomes:

0 0 1 1 1 1 0

1. ...

2

n

k n k k n k k n k k n k

k kP C C C C C C C C

Pr ' ' ' ' ' ' ' ' ' ' ' ' ' 1' ' ' ... ' ' ' ' ' ' ' 'ob O by A and K by B or i by A and K by B or or K by A and O by B

0 2 1 1 1 1

0 1 1

0

0

1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2

1 1 1...

2 2 2

k k n k k k n k k

k n k k n k

k k

k n k

k n k

k

P C C C C

C C

Therefore is equal to equation on (1)

Hence, it is proved


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28. Suppose that we want to generate a random variable that is equally likely to be either

0 or 1, and that all we have at our disposal is a biased coin that, when flipped, lands on

heads with some (unknown) probability p. Consider the following procedure:

1. Flip the coin, and let , either heads or tails, be the result.

2. Flip the coin again, and let be the result

3. If and are the same, retrun to step 1.

4. If is heads, set , otherwise set .

a) Show that the random variable generated by this procedure is equally likelyto be

either 0 or 1.

b) Could we use a simpler procedure that continues to flip the coin until the last two flips

are different, and then sets if the final flip is a head, and sets if it is a

tail?

Solutions :


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29. Soal ?

30. Let X be a poisson random variable with parameter λ. Show that { } increases

monotonically and then jecreases monotonicallyas i increases, teaching its maximum

when I is the largest integer not exceeding λ.

Hint : Consider { } { }

Jawab :

( ) { }

{ } { }


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{ }

( )

⁄

( )

{ } { }

⁄

( )

( )

( )

∫ ( ) ∫ ∫ ( )


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31. Persamaan variabel acak binomial dimana :

P(i) = ( ) pi

(1 – p)n – i

* ( ) =

( )

i = 0, 1, 2, … ,n

(i) P(x=2), saat n= 8, p=0,1

P(2) = ( ) (0,1)

2 (1 – 0,1)

8 – 2

=

( ) (0,01) (0,9)

6

= 28 (0,01) (0,531441)

= 0,1488 = 0,15

(ii) P(x=9), saat n= 10, p=0,95

P(9) = ( ) (0,95)

9 (1 – 0,95)

10 – 9

=

( ) (0,63025) (0,05)

= 10(0,63025) (005)

= 0,315125 = 0,32

(iii) P(x=0), saat n= 10, p=0,1

P(0) = ( ) (0,1)

0 (1 – 0,1)

10 – 0

=

( ) (1) (0,9)

10

= 1 (1) (0,9)10

= 0,34868 = 0,35

(iv) P(x=4), saat n= 9, p=0,2

P(4) = ( ) (0,2)

4 (1 – 0,2)

9 – 4

=

( ) (0,0016) (0,8)

5

= 126 (0,0016) (0,32768)

= 0,0661 = 0,07

32. Jika kamu membeli sebuah lotre dalam 50 lotre, kesempatan kamu mendapatkan hadiah

, berapa peluang kamu akan memenangkan hadiah :

a. paling sedikit satu kali

b. tepat satu kali

c. paling sedikit dua kali


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penyelesaian :

a. Paling sedikit satu kali

P { X 1} = 1 – P{ X = 0}

= 1 – ( ) (0,01)

0 (1 – 0,01)

50- 0

= 1 – (1) (1) (0,605)

= 1 – 0,605 = 0,395

b. Tepat satu kali

P { X 1} = ( ) (0,01)

1 (1 – 0,01)

50- 1

= (50) (0,01) (0,99)49

= (50) (0,01) (0,611)

= 0,3055 = 0,31

c. Paling sedikit dua kali

P { X 2} = 1 – (P{ X = 0} + P{ X = 1})

= 1 – ( ) (0,01)

0 (1 – 0,01)

50- 0 ( ) (0,01)

1 (1 – 0,01)

50- 1

= 1 – (1) (1) (0,605) – (50) (0,01) (0,611)

= 1 – 0,605 – 0,3055

= 1 – 0,91 = 0,09

33. Let X be a random variable with probability density

( ) { ( )

a). what is the value of c ?

b). what is the cumulative distribution function of X ?

jawab :

a)

∫ ( ) ∫ ∫ ( )

∫ ( )


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(

) (

)

b)

( ) ∫ ( )

∫

( )

( ) ∫ ( )

∫

( )

∫( )

[(

]

(

)

(

)

( )


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34. Let the probability density of X be given by

( ) { ( )

a) What is the value of c?

b) {

}

Answer :

a) ∫

∫ ( )

∫

∫

|

( ( )

( ) )

b) ∫ ( )

∫

( )

∫

|

(

(

)

(

)

) (

(

)

(

)

)


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35. The density of is given by

( ) {

What is the distribution of ? Find { }

Jawab :

Karena batas yang akan dicari berada didalam persamaan

maka diperoleh

∫

]

(

)

36. A point is uniformly within the disk of radius 1. That is, its density is

( )

Find the probability that its distance from the origin is less than

Jawab :

( )

∫ ( ) ( )

∫

( )

[ ]

[( ) ( )]

( )


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∫ ( ) ( )

∫

( )

[ ]

[( )]

∫ ( ) ( )

∫ ( ) ( )

37. Let 1 2, , ... , nX X X be independent random variables, each having a uniform distribution

over 0,1 . Let 1 2max , ,..., nM imum X X X . Show that the distribution of M, .mF , is

given by

, 0 1n

MF x x x

What is the probability density function of M ?

Solution

1

1

1

1

max ,...,

,...,

:

n

n

n

i

i

n

x

M

P M x P X X x

P X x X x

P X x

x

Functionof M

dF x P M x n

dx


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30

38. if the density function of M ?

( ) {

Answer :

{ } ∫

39. The random variable X has the following probability mass function

1 1 1(1) , (2) , (24)

2 3 6P P P . Calculate E(X) .

Jawab

Y = X Y = random variable

1(1) { 1}

2

1(2) { 2}

3

1(24) { 24}

6

y

y

y

P P Y

P P Y

P P Y

Jadi 1 1 1 31

( ) ( ) 1 2 24 5,172 3 6 6

E X E Y


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31

40. Suppose that two teams are playing a series of games, each of which is independently

won by team A with probability p and by team B with probability 1-p. The winner of the

serries is the first team to win 4 games. Find the expected number of games that are

played, and evaluate this quantity when 1

2p .

Solution :

[ ] ∑ ( )

( ) ( )

[ ] ( ) ( )

(

) (

)


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42. Suppose that each coupon obtained is, independent of what has been previously obtained,

equally likely to be any of m different types. Find the expected number of coupons one needs

to obtain in order to have at least one of each type

Hind: Lex X be the number needed. It is useful to represent X by

∑

Where each Xi is a geometric random variable

Answer:

∑

∑

43. An urn contains n + m balls, of which n are red and m are black. They are withdrawn

from the urn, one at the time and without replacement. Let X be the number of red balls

removed before the first black balls is chosen. We are interested in determining E [X]. To

obtain this quantity, number the red balls from 1 to end. Now define the random variables Xi,i

= 1, ….. , n , by.

1,

0,i

if red ball i is takenbeforeanyblack ball is chosenX

otherwise


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34

a. Express X in term of the X

1

n

i

i

x X

b. Find E X

1

1

1

1

1

i i

n

i

i

E X P X

P peluang terambil bola merahi sebelum semua nbola hitam

n

nE X E X

n

44. In exercise 43, let Y denote the number of red balls chosen after the first but before the

second black balls has been chosen.

a. Express Y as the sum of n random variables, each of which is equal to either 0 or 1.

b. Find E [Y].

c. Compare E [Y] to E[X] obtained in exercise 43.

d. Can you explain the result obtained in part (c) ?

45. A total of r keys are to be put, one at a time, in k boxes, with each key independently

being put in box with probability ∑ . Eachtime a key is put

in a nonempety box, we say that a collision occurs. Find the expected number of

collisions.

Answer :


5F © Mathematics

35

Let X be the number of collisions the X can be respresented as X = X1 + X2 + X3 + … +

Xk , where

Total number of keys = r

Each key is independently put in box I with probability Pi (i = 1, 2, 3, 4,…,k)

So probability the key is put in box 1 = P1 (for each of keys 1, 2, 3,…,r)

So probability that a key is not put in box 1 = 1 – P1 (for each of keys 1, 2, 3,…,r)

So,

PXiP ]1[ {there is at least one key in box 1}

= P1 {none of the 1, 2, 3,…,r keys are in box 1)

= )1(...)1()1()1[(1 1111 PPPP (r times)]

= rP )1(1 1

}0{0}1{1][ 1 ii XPXPXE

= rP )1(1 1

Maka: ][XE ][ 1XE + ][ 2XE + … + ][ kXE

1 ; if there is at least one key in

the box

0 ; otherwise

Xi


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36

= rP )1(1 1 + rP )1(1 2 + … + r

kP )1(1

= k – [ rP )1(1 1 + rP )1(1 2 + … + r

kP )1(1 ]

46. Consider three trials, each of which is either a success or not. Let denote the number of

success. Suppose that

a) What is the largest possible value of ?

b) What is the smallest possible value of ?

Let be 1 if trial is a success and 0 otherwise.

ANSWER

a) The largest value is 0,6 if then

1,8 = E [X] = 3 E [ ] = 3 P{

And also P {X = 3} = P { . That this is the largest value is seen

by Markov’s inequality which yields that

P {X

b) The smallest value is 0. To construct a probability scenario for which P{X = 3}= 0,

let U be a uniform random variable on (0, 1) and define

x1 = {

x2 = {

x3 {

47. If X is uniformly distributed over calculate .

{ }


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37

∫

∫ (

)

∫

|

48. SOAL ?

Solution :

Bukti bahwa 22E X E X .

2

2

2

2 2 2 2

2 2 2

22

(0) 0,2

(1) 0,5

(2) 0,3

,

(0) 0 0,2

(1) 1 0,5

(4) 2 0,3

:

0 (0,2) (1 )(0,5) (2 )(0,3) 1,7

0(0,2) 1(0,5) 2(0,3) 1,1 1,21

,

( )

1,7 1,21

Y

Y

Y

p

p

p

maka

P P Y

P P Y

P P Y

Jadi

E X

E X

Jadi

E x E x


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Memiliki kesamaan nilai, jika x = 0

22

22

( )

0 (0)

0 0

E x E x

E E

49. Let c be a constant. Show that.

ANSWER


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50. a coin, having probability pof landing heads, is flipped until the head appears for the rth

times. Let N denote the number of flips required. Calculate E

Answer:

Example:

∑

∑

∑ (

)

∑

Where;

∑


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51. Calculate the variance of the Bernaoulli random variabel

Answer:

Let ‘N’ denotes the number of flips repaired. Where N can be represented as N = N1 + N2

+….. + Nr.

Where Ni is the geometric random variables which denotes the number of flips required to

get i head.

1

1 2

1

1

1

1

...

(1 )

(1 )

1

1 1...

r

n

n

n

n

E N E N E N E N

E N np p

P n p

p

E N r timesP P

rE N

p


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52)

b

11

2 3

1 1

11

2 3

11

3

3 3

3 3

22

3

6 2

3

4

3

41

3

4 3

3

4

3 3 1(1 )

4 4 3

3 1 11 1

4 3 3

3 1 11 1

4 3 3

3 22

4 3

2 1

3 2

21

2

cc cx dx cx x

c cc c

c cc c

cc

c c

c

c

c

c

x dx x x


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42

c)

2

2 3

0

22

3

168

3

24 16

3

8

3

81

3

8 3

3

8

8

3

8 31

3 8

c x x dx

c

c

c

c

c

c

c

53. If X is uniform over 0, 1 , calculate nE X and nVar X .

: 0

,

,

n

x p xn

n

x p x jika X adalah diskrit

E X

x f x dx jika X adalah kontinu

1

0

1

0

11 1 1 1

0

1

1 0

1 0 1 1

1 1 1 1 1

n n

n

n n n n

E X x dx

x dx

x

n n n n n

Misal , nE X , maka


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2

2 2

2 2

2 2

2 2

2 2

22

2

2

2

2

n n

n n

n n

n n

n

n

n n

Var X E X

E X X

x x f x dx

x f x dx x f x dx f x dx

E X

E X

E X E X

Untuk nilai 2nE X :

12 2

0

12 1 2 1

0

1

2 1 2 1

n n

n n

E X x dx

x

n n

Jadi,

22 11 1

2 1 1

nnVar X

n n

54. Let X dan Y each take on either the value 1 or -1

Let

( ) { }

( ) { }

( ) { }

( ) { }

Suppose that [ ] [ ] . Show that

a. ( ) ( )

b. ( ) ( )

Let ( ). Find

c. Var ( ) [( ) ] [ ( )]


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d. Var ( ) [( ) ] [ ( )]

e. Cov ( ) [( ) ( ) ( )]

Answer:

a. ( ) ( )

( ) ( ) ∫ ∫ ( ) ∬ ( )

( ) ( ) ∫ ∫ ( )

∬ ( )

Jadi ( ) ( )

b. Karena [ ( ) ( )] [ ( ) ( )]

Maka [ ] [ ]

( ) ( )

55. Let X be a positive random variable having density function f(x). If ( )f x c for all x,

show that, for a > 0.

{ } 1P X a ac

Solutions :

[ ] ∫ ( )

∫ ( ) ∫ ( )

∫ ( )

∫ ( ) ∫ ( )

[ ] ∫ ( )

∫ ( ) [ ]

Since f(X)

Therefore,


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[ ] ∫

[ ] c( )

[ ]

Or [ ]

Hancew proved

56. Calculate, without using moment generating functions the variance of a binomial random

variable with parameters n and p.

Solutions :

If is binomial with parameters n dan p, then

[ ] ∑

( ) ( )

Writing ( )

[ ] ∑ ( )

( ) ( ) [ ]

∑

( )( )

( ) [ ]

( ) ∑( )

( ) ( )

( ) [ ]

( ) ∑ (

)

( ) [ ] (by j = i – 2)

( ) [ ( )]n-2 + E[X]

( ) [ ]

because E[X] = np, we arrive at

( ) [ ] ( [ ])

( )

( )


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58. Suppose that X and Y are independent countinous random variables.

Show that P{X ≤ Y}= { | } ( )yP X Y Y y f y dy

= { | } ( )xP X y Y y f y dy

= { } ( )yP X y f y dy

= ( ) ( )x xF y f y dy

The proof P{X≤Y} = ( ) ( )x xF y f y dy

59. soal ?

given that { }

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

60. Calculate the moment generating function of the uniform distribution on (0,1). Obtain

E[X] and Var [X] by differentiating.

ANSWER

E[X] dan Var [X]

(a,b) → (0,1)

( )

( )

( )

( )


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61. Suppose that X takes on each of the values 1,2,3 with probability 1

3, what is the moment

generating function ? Derive E[X], 2

[ ]E X and 3[ ]E X by differentiating the moment

generating function and the then compare the obtained result with a direct derivation of

these moments.

Solutions :

Given that

( )

( )

( )

By the definitionof moment generating function, we have,

( ) ∑ ( )

( )

Now

( )

Put

( )

[ ]

( )

[ ] ( )

( )

[ ] ( ) ( )


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49

Now

[ ] ∑ ( )

Which is same as calculated by ( )

[ ] ∑ ( )

[ ] ∑ ( )

Hence [ ] and [ ] are also same as calculated by

( ) ( )

62. Suppose the density of X is given by

( ) = {

Calculate the moment generating function, E[x] and Var (X)

Answer :


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50

Turunanpertama

( )

Misal :u =

v =

u' =

v' =

( )

(

)

( )

( )

(

)

Turunankedua

( )

(

)

Misal : u =

v =

u' =

v' =

( )

(

)

(

)

( )

( )


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51

Ambil : x = 1

( )

(

)

( )

( ) (

( ))

( )

(

)

( )

Ambil : x = 1

( )

( )

( )

( )

( )

Var(X) = E[ ] ( [ ])

=

(

)

=

=

(

)

=

(

)


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52

64

1 1

0 0

1 2

0

0

1

0

( ) ( 1)!

:

lim

lim ( 1)

( ) ( 1)

( 1) ( 1)

M

x n x n

M

M

x n M x n

M

M

x n

n n

bukti

e x dx e x dx

e x n e x dx

n n e x dx

n n


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53

66. Use Chebyshev’s inequality to prove the weak law of large numbers. Namely, if

are independent and identically distributed with mean and variance then, for any ,

{|

| }

Answer:

The weak law of large numbers (cf. the strong law of large numbers) is a result in probability

theory also known as Bernoulli's theorem. Let , ..., be a sequence of independent and

identically distributed random variables, each having a mean and standard deviation

. Define a new variable

(1)

Then, as , the sample mean equals the population mean of each variable.

(2)

(3)

(4)

(5)

In addition,

(6)

(7)

(8)

(9)

Therefore, by the Chebyshev inequality, for all ,

(10)

As , it then follows that

http://mathworld.wolfram.com/StrongLawofLargeNumbers.html

http://mathworld.wolfram.com/BernoullisTheorem.html

http://mathworld.wolfram.com/Mean.html

http://mathworld.wolfram.com/StandardDeviation.html

http://mathworld.wolfram.com/Mean.html

http://mathworld.wolfram.com/ChebyshevInequality.html


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(11)

(Khinchin 1929). Stated another way, the probability that the average |(

) |

for an arbitrary positive quantity approaches 1 as

67. Suppose that X is a random variable with mean 10 and variance 15. What can we say

about P{5<X<15} ?

Solutions :

Given that X is a random Variables with mean ( ) = 10 and variance ( ) = 15

We know that, Chebychev’s inequality is

P {| | } (

)

Also we can write

P { } = P{| | }

68. Let X1. X2, …,X10 be independent Poisson random variables with mean 1.

(i) Use the Markov Inequality to get a bound on P {X1 + … + X10 ≥ 15}

Contoh soal halaman 75

Markov Inequality

P {X > a} ≤ [ ]

{ } { }

≤

(ii) Use the central limit theorem to approximate P{X1 + … + X10 ≥ 15}

P{X = 15} = P{14,5 < X < 15,5}

http://mathworld.wolfram.com/Positive.html


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{

√

√

√ }

{

√ }

( ) ( )

( ) { ( ) } ( )

P{X = 15} ≈ 2Ф(0,18) – 1

P{X = 15} ≈ 0,1428

70. Show that

∑

Hint: Let Xn Poisson with mean n. Use the central limit theorem to show that P { Xn n }

.

Answer:

{∑

} ∑

But for n large ∑ has approximately a normal distribution with mean 0, and so

the result follows.

71. Let X denote the number of white balls selected when k balls are chosen at random from

an urn containing n white and m black balls.

(i) Compute P {X=i}

(ii) Let, for i = 1, 2, ..., k; j=1, 2, ..., n,

Xi = {


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56

Yj ={

Compute E [X] in two ways by expressing X first as a function of the Xi’s and then of

theYj’s

Answer:

X = X1 + X2 + X3 + ... Xn

Xi = {

P { Xi = 1} = P {i th ball selected is white} =

So, E [ Xi] = 1P {Xi = 1}+ 0P {Xi = 0}=

Y = Y1 + Y2 + Y3 + ... Yn

Yj = {

P { Yj = 1} = P {j th white ball is selected} =

So, E [ Yj] = 1P {Yj = 1}+ 0P {Xj = 0}=

72. show that Var(x)= 1 when X is the number of men that select their own hats in example

2.31

For the matching problem, letting X = X1 + … + XN, where

Xi = {

We obtain

( ) ∑ ( )

∑∑ ( )

ifith man selects his own hat

otherwise


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Since P{ }

( )

(

)

Also

( ) [ ] [ ] [ ]

Now,

XiXj= {

And thus

[ ] { } { } { | }

Hence,

( )

( ) (

)

( )

And

( )

(

)

( )

If theith and jth men both selects their own hats

otherwise


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75. Let < ... < denote a set of n numbers, and consider any permutation of

these numbers. We say that there is an inversion of and in the permutation if

and precedes . For instance the permutation 4, 2, 1, 5, 3 has 5 inversions

(4,2), (4,1), (4,3), (2,1), (5,3). Consider now a random permutation of ,

in the sense that each of the n! Permutation is equally likely to be chosen and let N

denote the number of inversions in this permutation. Also, let

N! = number of precedes in the permutation

And note that N = ∑

(i) Show that N1,..., Nn are independent random variables.

(ii) What is the distribution of N1

(iii) Compute E[ N ] and Var (N).

Solutions :

a) knowing the value of ..., is equivalent to knowing the relative ordering of the

elements ..., . For instance, if , then in random

permutation is before which is before . The independence result follows for

clearly the number of ..., that follow does not probabilistically depend on

the relative ordering of ...,

b) { }

which follows since of elements ..., the element is equally likely to be

first or second or...or ( )

76. Let and be independent random variables with means and and variances

and . Show that ( )

Answer:

( ) ( ) ( )

( ) ((

)

( ) )


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( ) ((

) ) ((

)

( ) )

( )

( ) (((

) )

) ((

)

( ) )

( ) (( )

)

Maka dari keterangan di atas didapat:

[ ] ( )

[( ) ] ( )

( ) [( ) ] ( [ ])

( ) ( )

( )

( )

( )

77. Let and be independent normal random variables each having parameters and

. Show that is independent of .

Solution :

Given that X and Y be two independent Normal variables, each having the parameters

and 2

Mx (t) =

= My (t)

Also Mx+y (t) = Mx(t) x My (t)

78. Let ( ) denote the joint moment generating function of .

a) Explain how the moment generating function , ( ) can be obtained from

( )


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61

b) Show that are independent if and only if ( ) ( )

( )

Diketahui:

( ) [ ]

( ) ( ) ( )

( ) ( ) ( ) ( ) ...1

a) Untuk mendpatkan fungsi momen generasi , ( ) maka dari persamaan …1

didapat

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

Dari persamaan di atas didapat untuk fungsi momen generasi adalah ( )

b) Untuk menunjukkan adalah independen jika dan hanya jika

( )

( )

( ) maka dari persamaan …1 didapat

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

Solution Manual for Introduction to Probability Models 7th

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