Post on 01-Jan-2016
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
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}
Solution Manual for : Introduction to Probability Models
5F © Mathematics
2
‘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)} =
Solution Manual for : Introduction to Probability Models
5F © Mathematics
3
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)} =
Solution Manual for : Introduction to Probability Models
5F © Mathematics
4
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)} =
Solution Manual for : Introduction to Probability Models
5F © Mathematics
5
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
6
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
7
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:
Solution Manual for : Introduction to Probability Models
5F © Mathematics
8
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
9
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
( ) ( ) (
)
(
)
Solution Manual for : Introduction to Probability Models
5F © Mathematics
10
( ) (
) (
)
( )
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
11
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) ( )
( )
( )
( )
Solution Manual for : Introduction to Probability Models
5F © Mathematics
12
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
13
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: ?
Solution Manual for : Introduction to Probability Models
5F © Mathematics
14
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% >> ( ) ( ) (
)
(
)
(
) (
)
Solution Manual for : Introduction to Probability Models
5F © Mathematics
15
Solution Manual for : Introduction to Probability Models
5F © Mathematics
16
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
17
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 = ( ) ( )
Solution Manual for : Introduction to Probability Models
5F © Mathematics
18
= ( ) ( )
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 → ( ) {( )}
[ ]
Solution Manual for : Introduction to Probability Models
5F © Mathematics
19
( ( ) )
( ( ) {(
) } ( ) {(
) })
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
20
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
21
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 :
Solution Manual for : Introduction to Probability Models
5F © Mathematics
22
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 :
( ) { }
{ } { }
Solution Manual for : Introduction to Probability Models
5F © Mathematics
23
{ }
( )
⁄
( )
{ } { }
⁄
( )
( )
( )
∫ ( ) ∫ ∫ ( )
Solution Manual for : Introduction to Probability Models
5F © Mathematics
24
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
25
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)
∫ ( ) ∫ ∫ ( )
∫ ( )
Solution Manual for : Introduction to Probability Models
5F © Mathematics
26
(
) (
)
b)
( ) ∫ ( )
∫
( )
( ) ∫ ( )
∫
( )
∫( )
[(
]
(
)
(
)
( )
Solution Manual for : Introduction to Probability Models
5F © Mathematics
27
34. Let the probability density of X be given by
( ) { ( )
a) What is the value of c?
b) {
}
Answer :
a) ∫
∫ ( )
∫
∫
|
( ( )
( ) )
b) ∫ ( )
∫
( )
∫
|
(
(
)
(
)
) (
(
)
(
)
)
Solution Manual for : Introduction to Probability Models
5F © Mathematics
28
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 :
( )
∫ ( ) ( )
∫
( )
[ ]
[( ) ( )]
( )
Solution Manual for : Introduction to Probability Models
5F © Mathematics
29
∫ ( ) ( )
∫
( )
[ ]
[( )]
∫ ( ) ( )
∫ ( ) ( )
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
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 :
[ ] ∑ ( )
( ) ( )
[ ] ( ) ( )
(
) (
)
Solution Manual for : Introduction to Probability Models
5F © Mathematics
32
Solution Manual for : Introduction to Probability Models
5F © Mathematics
33
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
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 :
Solution Manual for : Introduction to Probability Models
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
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 .
{ }
Solution Manual for : Introduction to Probability Models
5F © Mathematics
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
38
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
39
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;
∑
Solution Manual for : Introduction to Probability Models
5F © Mathematics
40
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
41
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
43
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 ( ) [( ) ] [ ( )]
Solution Manual for : Introduction to Probability Models
5F © Mathematics
44
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,
Solution Manual for : Introduction to Probability Models
5F © Mathematics
45
[ ] ∫
[ ] 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
( ) [ ] ( [ ])
( )
( )
Solution Manual for : Introduction to Probability Models
5F © Mathematics
46
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)
( )
( )
( )
( )
Solution Manual for : Introduction to Probability Models
5F © Mathematics
47
( )
( )
( ) [ ] ( )
Solution Manual for : Introduction to Probability Models
5F © Mathematics
48
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
( )
[ ]
( )
[ ] ( )
( )
[ ] ( ) ( )
Solution Manual for : Introduction to Probability Models
5F © Mathematics
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 :
Solution Manual for : Introduction to Probability Models
5F © Mathematics
50
Turunanpertama
( )
Misal :u =
v =
u' =
v' =
( )
(
)
( )
( )
(
)
Turunankedua
( )
(
)
Misal : u =
v =
u' =
v' =
( )
(
)
(
)
( )
( )
Solution Manual for : Introduction to Probability Models
5F © Mathematics
51
Ambil : x = 1
( )
(
)
( )
( ) (
( ))
( )
(
)
( )
Ambil : x = 1
( )
( )
( )
( )
( )
Var(X) = E[ ] ( [ ])
=
(
)
=
=
(
)
=
(
)
Solution Manual for : Introduction to Probability Models
5F © Mathematics
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
54
(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}
Solution Manual for : Introduction to Probability Models
5F © Mathematics
55
{
√
√
√ }
{
√ }
( ) ( )
( ) { ( ) } ( )
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 = {
Solution Manual for : Introduction to Probability Models
5F © Mathematics
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
Solution Manual for : Introduction to Probability Models
5F © Mathematics
57
Since P{ }
( )
(
)
Also
( ) [ ] [ ] [ ]
Now,
XiXj= {
And thus
[ ] { } { } { | }
Hence,
( )
( ) (
)
( )
And
( )
(
)
( )
If theith and jth men both selects their own hats
otherwise
Solution Manual for : Introduction to Probability Models
5F © Mathematics
58
Solution Manual for : Introduction to Probability Models
5F © Mathematics
59
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:
( ) ( ) ( )
( ) ((
)
( ) )
Solution Manual for : Introduction to Probability Models
5F © Mathematics
60
( ) ((
) ) ((
)
( ) )
( )
( ) (((
) )
) ((
)
( ) )
( ) (( )
)
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
( )
Solution Manual for : Introduction to Probability Models
5F © Mathematics
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
( ) ( ) ( ) ( )
( ) ( )
( ) ( )