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Probability, Third Edition
By David J Carr & Michael A Gauger
Published by BPP Professional Education
Solution 4.1
(i)
The sample space is:
S ={,,,,,, ,
BBBR BBRB BBRR BRBB BRBR BRRB BRRR RBBB ,
,,,,,,
RBBR RBRB RBRR RRBB RRBR RRRB RRRR }
(ii)
The random variable X is defined as:
X BBBR ( X BRBB (
() =X BBRB ) =() =X RBBB ) =1
() =X BRBR ) =X BRRB ) =X RBBR ) =X RBRB ) =()
X BBRR (((( X RRBB =2
() =X RBRR ) =X RRBR ) =X RRRB )
X BRRR ((( =3
() =
X RRRR 4
(iii)
The random variable Y is defined as:
() =X BRBR ) =() =X RBBR ) =() =X RRBB )
X BBRR ( X BRRB ( X RBRB ( =0
X BBBR ( X BRBB (( ( ( (
() =X BBRB ) =() =X RBBB ) =X BRRR ) =X RBRR ) =X RRBR ) =X RRRB ) =2
() =
X RRRR 4
Solution 4.2
The sample space is S ={(, ,):70 l 120 ,0 w 12 ,0 t 4
lwt <<
<<
<<
}, where l is the length of a plank, w is the
Solutions to practice questions h
Chapter 4
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width of the plank, and t is the thickness of the plank.
The random variable X is a continuous random variable defined as:
(, ,)=lwt for all (, ,).S
Xlwt lwt
Solution 4.3
Since the probabilities must sum to 1, we have:
4
3333
Err:511
()
()
()
+3a +4a =100a3
1 ia +2a
i=1
31
.a =.
a =0.2154
100
1
Solution 4.4
Since all probabilities must lie between zero and one, we have:
Err:509
4
.=
.==
11 13
Err:509
4
.=
.==
1 3 1
Err:509
4
.=
.==
11
2
3
Probability Solutions to practice questions h
Chapter 4
Sheet1
Page 3
2
Err:509
4
.=
.==
31
4
1
4
The only range which satisfies all of these is:
.==1
3
1
4
Solution 4.5
The following table shows the product of the scores on the two dice.
Score on die 1
1 2 3 4 5 6
Score on die 2
1 1 2 3 4 5 6
2 2 4 6 8 10 12
3 3 6 9 12 15 18
4 4 8 12 16 20 24
5 5 10 15 20 25 30
6 6 12 18 24 30 36
By counting the number of (equally likely) outcomes, the probability distribution is:
12 2
Err:520
Pr( X ==
Pr( X =18) =
Pr( X 1) 8)
3636 36
21 2
Err:520
Pr( X ==
Pr( X =20) =
Pr( X 2) 9)
36 36 36
222
Pr( X ==
Pr( X =10) =
Pr( X =
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3) 24) =
3636 36
341
Pr( X ==
Pr( X =12) =
Pr( X =
4) 25) =
3636 36
22 2
Pr( X ==
Pr( X =15) =
Pr( X =
5) 30) =
3636 36
411
Pr( X ==
Pr( X =
36)
6) 16) =
Pr( X ==
3636 36
3
Solution 4.6
Let N be the random number of claims filed in the 3year period. Then N is a discrete random variable with
possible values 0,1,2,.
.. The first step is to compute the probability function pn =Pr (Nn).
Err:520
The recursive relation pn+1 =0.2
pn leads to the following:
pn =0.2
pn1 =0.2 2
pn2 "
0.2n
0
h
BPP Professional Education
Solutions to practice questions h
Chapter 4 Probability
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Page 5
Err:520
p
But since all probabilities must add to 1:
1 =pp1 "
p (1 +0.2 +0.2 2 +.
.)=p0
1 .p =0.8 geometric series )
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This note was uploaded on 09/08/2008 for the course STT 490 taught by Professor Dudziak during the Spring '08 term at Michigan State University.
 Spring '08
 Dudziak

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