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 5F © Mathematics

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 : elements that are 45 elements. It’s

The sample space of this experiment will contain 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}

1


‘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)} =

2


Step 2 x p (x)

1

2

3

4

5

6

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 p (x)

1

2

3

4

5

6

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)} =

3


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 p (x)

2

3

4

5

6

7

8

9

10

11

12

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)} =

4


6. Suppose five fair coins are tossed. Let E be the event that all coins land heads. Define the random variable I E 1, if E occurs IE = 0, if E c occurs For what outcomes in the original sample space does I E equal 1 ? What is P I E  1 ? 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

5


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

1, Jadi ,

32

IE 

dan 31

0

,

=

32

  P 1  P  0 

1 31 32   1 32 32 32

6


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:

P(0)  (0,3)3  0, 027 P(1)  3(0,3) 2 (0, 7)  0,189 P(2)  3(0,3)(0, 7) 2  0, 441 P(3)  (0, 7)3  0,343 P(0)  P(1)  P(2)  P(3)  0, 027  0,189  0, 441  0,343  1

8. SOAL ? Solution : 1

1 1 1  0 2dx   2 x  0  2 1



1dx   x



1

1

0

Jadi P( X ) 

1 3 1  2 2

9. If the distribution function of F is given by b0 0, 1  , 0  b 1 2 3 1 b  2  , 5 F b   4 , 2b3 5 9  , 3  b  3,5 10 1, b  3,5 Calculate the probability mass function of X Solution:

7


1 0 2 3 1 P 1   5 2 4 3 P  2   5 5 9 4 P  3   10 5 9 P  3.5   1  10 P  0



1 2 1  10 1  5 1  10 1  10 

10. SOAL ? Solution : 1 1 1 1 . .  6 6 6 36 6 6 6 P(paling banyak 6) = . .  1 6 6 6 1 1  P(paling banyak 1 dan 6) = .1  36 36

P(paling banyak 1) =

8


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 Diketahui

i2

:

n4 p

1 2

n n i p  i     pi 1  p  i  p  2

 4 1       2 2 

2

 1 1    2

42

 4 1   1  3         2 2   2  8 2

2

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 ( )

( )( ) (

)

9


( )

(

)( )

( ) 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 

 P i  i 0

10   1      i 7  i   2  10

10

14. SOAL ?

6 1   1  P( x  3)        3 2   2  6!  1   1   .    3!3!  8   8  6.5.4.3!  3!.3.2.1 1  20. 64 20 5   64 16 3

3

10


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 (

( )

)

(

)

(

)

(

)

(

) (

(

)

(

Mode is the value of a. in case, ( Let ( (

)

)

)

(

)

(

)

(

)

()

for which ( ) is maximum. Then there are two cases arrises :

) is an integer )

( )

and , then

(an integer)

( ) ( )

11


from ( ) it obvious that ( ) (

)

Hence if (

)is integer ( ) then (

(

)

b. in case, ( Let (

(

) inreases till

and

) and after that it begins to decreases.

) is not an integer )

+y (where and

is an integer

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 12


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 x 1 times, the second x2 times, and the rthxrtimes is n! p1x p2 when x1  x2  ...  xr  n x1 ! x2 !...xr ! 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: ?

13


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 x1  x2  ...  xk ? 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 20% >> ( )

( )( ) ( )

( )(

)

For 30% >> ( )

( )( ) ( )

( )(

)

For 50% >>

(

)

14


15


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 ? 5

1 1 P(H pada lemparan kelima) =     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  n  1 r nr P {X=n} =   p 1  p  , n  r  r 1 

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

16


P( X  n)  P.P ( X  n  1)  n  1 n 1 n 1( r 1) P ( X  n)  P   P (1  p )  r 1   n  1 r 1 n 1 r 1   P.P (1  p)  r 1   n  1 r nr P ( X  n)   P (1  p )  r 1 

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 7 th 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= ( )

(

)

17


= ( )(

)

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 → (

)

)

{( ) {(

} )}

[ ]

18


(

(

) ) (

(

)

{( )

(

}

)

{( )

[ ]

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 →  B wins 3 games out of 4 → (

)

{( )

) (

 A wins 3 games out of 5 →  B 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 19


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

k

0

nk

1  C1k    2

0

k

0

nk

1 C    2

1 1 1 1 P  C0k     .C0nk     2 2  2  2 1 1 1 1 P  C     .C0nk     2 2  2  2 k 0

k 1

k

1

1    2

n  k 1

k

1

1   2

n  k 1

1

 1  nk  1    .C1    2  2

1

 1  nk  1    .C1   2 2

k

1 1  ...  CKk   .Ckn k    2  2

n k

( )

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

1  P    C0k .C0n k  C1k C1n k  ...  Ckk Ckn k  2

Writing the above sequence in reverse order. n

1 P    Ckk .C0nk  Ckk1Ckn1k  ...  C1k C1n k  C0k C0n k  2

n  n  Also we know that,   .   r  nr

Therefore, the above equation becomes: n

1 P    C0k .C0nk  C1k C1nk  ...  Ckk1C1n k  Ckk C0nk  2 Pr ob ' O ' by ' A ' and ' K ' by ' B ' or ' i ' by ' A ' and ' K 1' by ' B ' or...or ' K ' by ' A ' and ' O ' by ' B '

0

k

k

1 1 1 1 P  C     Ckn k     2 2 2 2 k 0

k

0

1 1 1 ...  C   C0n k     2 2 2

n2k

1

1 1 C     2 2 k 1

k 1 n k k 1

C

1   2

k 1

1   2

n  k  k 1



nk

k k

Therefore is equal to equation on (1) Hence, it is proved

20


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 3. If

and

4. If

is heads, set

be the result

are the same, retrun to step 1. , 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 :

21


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 : () {

{ }

} {

}

22


⁄ {

}

{

}

(

)

{

(

)

} ⁄ (

(

∫ ( )

)

∫

(

)

)

∫ (

)

23


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)

= = = =

(ii)

= = = =

(0,01) (0,9)6

)

28 (0,01) (0,531441) 0,1488 = 0,15

(

) (0,95)9 (1 – 0,95)10 – 9

(

)

(0,63025) (0,05)

10(0,63025) (005) 0,315125 = 0,32

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

= =

(iv)

(

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

(iii)

( ) (0,1)2 (1 – 0,1)8 – 2

(

) (0,1)0 (1 – 0,1)10 – 0

(

)

(1) (0,9)10

= 1 (1) (0,9)10 = 0,34868 = 0,35 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 24


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)

∫ ( )

∫ (

∫

∫ (

)

)

25


(

)

(

)

b)

( )

∫ ( )

∫ (

( )

)

∫ ( )

∫ (

∫( [( ( (

)

) ] ) )

( )

26


34. Let the probability density of X be given by ( )

{

(

)

a) What is the value of c? {

b)

}

Answer : a) ∫

(

∫

)

∫

∫ | (

b)

( )

∫ ∫

( ) )

(

) (

)

∫

| ( ( )

( ) )

( ( )

( ) )

27


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 :

(

) (

∫ ∫ [ [(

) ( (

) )

] )

(

)]

( )

28


∫

( ) ( )

∫

( )

[ ] [(

∫ (

)]

) (

)

∫ ( ) ( )

37. Let X1 , X 2 , ... , X n be independent random variables, each having a uniform distribution over  0,1 . Let M  max imum  X1 , X 2 ,..., X n  . Show that the distribution of M, Fm . , is given by

FM  x   x n ,

0  x 1

What is the probability density function of M ? Solution P M  x

 P max  X 1 ,..., X n   x  P  X 1  x,..., X n  x n

  P  X i  x i 1

 xn Function of M : FM  x  

d P M  x  n x 1 dx

29


38. if the density function of M ? ( )

{

Answer : {

39. The

}

random

∫ variable

X

has

the

following

probability

mass

function

1 1 1 P(1)  , P(2)  , P(24)  . Calculate E(X) . 2 3 6

Jawab Y = X  Y = random variable

Py  (1)  P{Y  1}  Py (2)  P{Y  2} 

1 2

1 3

Py (24)  P{Y  24} 

1 6

1 1  1  31 Jadi E ( X )  E (Y )  1   2    24     5,17  2  3 6 6

30


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 p 

1 . 2

Solution :

[ ]

∑

()

( ) [ ] ( )

( ) ( ) (

(

) )

31


32


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 X i,i = 1, ….. , n , by. 1, Xi   0,

if red ball i is takenbefore any black ball is chosen otherwise

33


a. Express X in term of the X n

x   Xi i 1

b. Find E  X  E  Xi 

 P  X i  1  P  peluang terambil bola merah i sebelum semua n bola hitam  n

1  n  1

E  X    E  Xi   i 1

n  n  1

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 :

34


Let X be the number of collisions the X can be respresented as X = X 1 + X2 + X3 + … + Xk , where 1

; if there is at least one key in the box

0

; otherwise

Xi

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, P[ Xi  1]  P {there is at least one key in box 1}

= 1  P {none of the 1, 2, 3,…,r keys are in box 1) = 1  [(1  P1 )  (1  P1 )  (1  P1 )  ...  (1  P1 ) (r times)] = 1  (1  P1 ) r

E[ X 1 ]  1  P{X i  1}  0  P{X i  0}

= 1  (1  P1 ) r

Maka: E[X ]  E[ X 1 ] + E[ X 2 ] + … + E[ X k ]

35


= 1  (1  P1 ) r + 1  (1  P2 ) r + … + 1  (1  Pk ) r = k – [ 1  (1  P1 ) r + 1  (1  P2 ) r + … + 1  (1  Pk ) r ]

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

then ] = 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

.

}

36


∫ ∫

(

)

∫ |

48. SOAL ? Solution : Bukti bahwa E  X 2    E  X  . 2

p (0)  0, 2 p (1)  0,5 p (2)  0,3 maka,

 P (1)  P Y  1   0,5 P (4)  P Y  2   0,3

PY (0)  P Y  02   0, 2 2

Y

2

Y

Jadi : E  X 2   02 (0, 2)  (12 )(0,5)  (22 )(0,3)  1, 7

 E  X   0(0, 2)  1(0,5)  2(0,3)  1,1 2

2

2

 1, 21

Jadi, E  x 2    E ( x) 

2

1, 7  1, 21

37


Memiliki kesamaan nilai, jika x = 0

E  x 2    E ( x)

2

E 02    E (0)

2

00

49. Let c be a constant. Show that.

ANSWER

38


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; ∑

39


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 = N 1 + N2 +….. + Nr. Where Ni is the geometric random variables which denotes the number of flips required to get i head. E  N   E  N1   E  N 2   ...  E  N r  

E  N1    np (1  p )

n1

n 1



 P  n (1  p ) n 1 n 1

1 p 1 1  E  N    ... r times P P 

EN 

r p

40


52) b 1

c 3  1 c  cx dx  cx  3 x  1 1

2

c  c  c    c   3  3 c c c c 3 3 2c  2c  3 6c  2 c  3 4c  3 4c 1 3 4c  3 c

3 4 1

3 3  1 3 2 1 4 (1  x )dx  4  x  3 x  1 1

3 1  1    1    1    4 3  3  3 1 1  1   1   4 3 3 3 2  2  4 3 2 1   3 2 2  1 2

41


c) 2

2   c  2 x 2  x 3 dx  3  0  16   c 8   3   24  16   c   3  8  c 3 8 c 1 3 8c  3 c

3 8

8  c 3 8 3  1 3 8

53. If X is uniform over  0, 1 , calculate E  X n  and Var  X n  .   xn p  x  ,   x: p x   0 E  X n       x n f  x  dx,  

jika X adalah diskrit jika X adalah kontinu

1 n x dx 0 1 0

E  X n   

1

1

  x n dx 0

1

x n 1 1n 1 0n 1 1n 1 1      n 1 0 n 1 n 1 n 1 n 1

Misal , E  X n    , maka 42

Solution Manual for : Introduction to Probability Models 5F © Mathematics 2 Var  X n   E  X n      

 E  X 2 n  2 X n   2  





x

2n

 2 x n    2  f  x  dx













  x 2 n f  x  dx  2  x n f  x  dx   2 

f  x  dx

 E  X 2 n   2    2  E  X 2 n    2



 E  X 2 n   E  X n 



2

Untuk nilai E  X 2n  : 1

E  X 2 n    x 2 n dx 0

1

x 2 n 1 12 n 1   2n  1 0 2n  1 Jadi, 12 n1  1  Var  X     2n  1  n  1 

2

n

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

)

{

}

(

)

{

}

(

)

{

}

(

)

{

Suppose that [ ] a. b. Let

( (

) (

c. Var ( )

[ ] (

)

}

(

. Show that ) )

). Find [( ) ]

[ ( )]

43


[( ) ] [ ( )] 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.

P{X  a}  1  ac Solutions :

[

]

( )

∫

( )

∫

( )

∫

( )

∫

( )

]

∫

( )

[

∫

∫

( )

[

∫

( )

]

Since f(X) Therefore, 44


[

]

∫

[

]

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 ( )

[

]

( [ ]) (

) (

)

45


58. Suppose that X and Y are independent countinous random variables. P{X ≤ Y}= 

Show that









=





=





=

The proof P{X≤Y} =







P{ X  Y | Y  y} f y ( y )dy







P{X  y | Y  y} f x ( y )dy

P{X  y} f y ( y )dy Fx ( y) f x ( y) dy

Fx ( y) f x ( 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) ( )

( )

(

)

(

)

46


( ) (

( )

[ ]

)

(

)

47


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

1 , what is the moment 3

2

generating function ? Derive E[X], E[ X ] and E[ X 3 ] 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 ( ) [ ] ( ) [

]

( )

( ) [

]

( )

(

) 48


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 :

49


Turunanpertama ( )

Misal :u =

v=

u' =

v' =

( )

(

)

( ) ( )

(

)

(

)

Turunankedua ( )

Misal : u =

v=

u' =

v' =

( )

(

)

(

)

( )

( )

50


Ambil : x = 1 ( )

(

( )

( )

( )

( )

)

(

( ))

( )

Ambil : x = 1 ( ) ( )

( )

( )

( )

Var(X) = E[

]

( [ ]) (

=

)

= =

(

=

(

) )

51


64

(n)  (n  1)! bukti : 

M

 x n 1  x n 1  e x dx  lim  e x dx M 

0

  lim  e  x x n 1  M  

0 M 0

M  (n  1)  e  x x n  2 dx   0

M

(n)  (n  1)  e  x x n 1dx 0

 (n  1)(n  1)

52


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 53


(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
) = 10 and variance (

) = 15

We know that, Chebychev’s inequality is P {|

|

}

(

)

} = P{|

|

Also we can write 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}

54


{

√

√

{

(

)

}

√ (

)

{ (

)

}

√

(

)

}

(

)

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 { X n n } . Answer: {∑ But for n large ∑ the result follows.

}

∑

has approximately a normal distribution with mean 0, and so

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 = {

55


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 ifith man selects his own hat

Xi = {

otherwise

We obtain

( )

∑

( )

∑∑

(

)

56


Since P{

} ( )

(

)

[

)

Also (

[ ] [ ]

]

Now, If theith and jth men both selects their own hats

XiXj= { otherwise

And thus

[

]

{

}

{

} {

|

}

Hence, (

)

(

( )

)

(

)

And ( )

( )

(

)

57


58


75.

Let

< ... <

denote a set of n numbers, and consider any permutation of

these numbers. We say that there is an inversion of and

precedes

and

in the permutation if

. 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 ) first or second or...or (

76. Let and

and

...,

the element

be independent random variables with means (

. Show that

is equally likely to be

and

and variances

)

Answer: ( )

( )

( )

(

(

( ) )

(

) )

59


( )

((

)

)

(

(

)

(

) )

(

)

( ) ( )

(((

( )

)

((

)

)

)

(

(

) )

)

Maka dari keterangan di atas didapat: [

]

[(

( ) ) ]

(

)

(

)

(

)

(

)

(

) [(

) ]

(

77. Let

and

( [

])

)

(

)

be independent normal random variables each having parameters

. Show that

is independent of

and

.

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

) 60


are independent if and only if (

b) Show that

)

( )

( )

Diketahui: (

)

(

)

(

)

[

] ( (

)

)

(

(

)

)

(

) ...1

a) Untuk mendpatkan fungsi momen generasi

,

( ) maka dari persamaan …1

didapat (

)

(

)

(

)

( )

(

)

( )

(

)

( )

Dari persamaan di atas didapat untuk fungsi momen generasi

adalah

( )

adalah independen jika dan hanya jika

b) Untuk menunjukkan (

)

( )

(

)

(

(

)

( )

( ) maka dari persamaan …1 didapat )

(

( )

)

(

)

( )

61

Solution Manual for Introduction to Probability Models 7th

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