p chapter 4 solutions 3rd fall 2007

# p chapter 4 solutions 3rd fall 2007 - Probability, Third...

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Sheet1 Page 1 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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Sheet1 Page 2 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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Sheet1 Page 4 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 3-year 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 pn-1 =0.2 2 pn-2 " 0.2n 0 h BPP Professional Education Solutions to practice questions h Chapter 4 Probability
Sheet1 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.

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p chapter 4 solutions 3rd fall 2007 - Probability, Third...

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