Chapter5

# Chapter5 - CHAPTER 5 Section 5.1 1. a. b. c. P(X = 1, Y =...

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175 CHAPTER 5 Section 5.1 1. a. P(X = 1, Y = 1) = p(1,1) = .20 b. P(X 1 and Y 1) = p(0,0) + p(0,1) + p(1,0) + p(1,1) = .42 c. At least one hose is in use at both islands. P(X 0 and Y 0) = p(1,1) + p(1,2) + p(2,1) + p(2,2) = .70 d. By summing row probabilities, p x (x) = .16, .34, .50 for x = 0, 1, 2, and by summing column probabilities, p y (y) = .24, .38, .38 for y = 0, 1, 2. P(X 1) = p x (0) + p x (1) = .50 e. P(0,0) = .10, but p x (0) p y (0) = (.16)(.24) = .0384 .10, so X and Y are not independent. 2. a. y p(x,y) 0 1 2 3 4 0 .30 .05 .025 .025 .10 .5 x 1 .18 .03 .015 .015 .06 .3 2 .12 .02 .01 .01 .04 .2 .6 .1 .05 .05 .2 b. P(X 1 and Y 1) = p(0,0) + p(0,1) + p(1,0) + p(1,1) = .56 = (.8)(.7) = P(X 1) P(Y 1) c. P( X + Y = 0) = P(X = 0 and Y = 0) = p(0,0) = .30 d. P(X + Y 1) = p(0,0) + p(0,1) + p(1,0) = .53 3. a. p(1,1) = .15, the entry in the 1 st row and 1 st column of the joint probability table. b. P( X 1 = X 2 ) = p(0,0) + p(1,1) + p(2,2) + p(3,3) = .08+.15+.10+.07 = .40 c. A = { (x 1 , x 2 ): x 1 2 + x 2 } { (x 1 , x 2 ): x 2 2 + x 1 } P(A) = p(2,0) + p(3,0) + p(4,0) + p(3,1) + p(4,1) + p(4,2) + p(0,2) + p(0,3) + p(1,3) =.22 d. P( exactly 4) = p(1,3) + p(2,2) + p(3,1) + p(4,0) = .17 P(at least 4) = P(exactly 4) + p(4,1) + p(4,2) + p(4,3) + p(3,2) + p(3,3) + p(2,3)=.46

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Chapter 5: Joint Probability Distributions and Random Samples 176 4. a. P 1 (0) = P(X 1 = 0) = p(0,0) + p(0,1) + p(0,2) + p(0,3) = .19 P 1 (1) = P(X 1 = 1) = p(1,0) + p(1,1) + p(1,2) + p(1,3) = .30, etc. x 1 0 1 2 3 4 p 1 (x 1 ) .19 .30 .25 .14 .12 b. P 2 (0) = P(X 2 = 0) = p(0,0) + p(1,0) + p(2,0) + p(3,0) + p(4,0) = .19, etc x 2 0 1 2 3 p 2 (x 2 ) .19 .30 .28 .23 c. p(4,0) = 0, yet p 1 (4) = .12 > 0 and p 2 (0) = .19 > 0 , so p(x 1 , x 2 ) p 1 (x 1 ) p 2 (x 2 ) for every (x 1 , x 2 ), and the two variables are not independent. 5. a. P(X = 3, Y = 3) = P(3 customers, each with 1 package) = P( each has 1 package | 3 customers) P(3 customers) = (.6) 3 (.25) = .054 b. P(X = 4, Y = 11) = P(total of 11 packages | 4 customers) P(4 customers) Given that there are 4 customers, there are 4 different ways to have a total of 11 packages: 3, 3, 3, 2 or 3, 3, 2, 3 or 3, 2, 3 ,3 or 2, 3, 3, 3. Each way has probability (.1) 3 (.3), so p(4, 11) = 4(.1) 3 (.3)(.15) = .00018 6. a. p(4,2) = P( Y = 2 | X = 4) P(X = 4) = 0518 . ) 15 (. ) 4 (. ) 6 (. 2 4 2 2 = b. P(X = Y) = p(0,0) + p(1,1) + p(2,2) + p(3,3) + p(4,4) = .1+(.2)(.6) + (.3)(.6) 2 + (.25)(.6) 3 + (.15)(.6) 4 = .4014
Chapter 5: Joint Probability Distributions and Random Samples 177 c. p(x,y) = 0 unless y = 0, 1, …, x; x = 0, 1, 2, 3, 4. For any such pair, p(x,y) = P(Y = y | X = x) P(X = x) = ) ( ) 4 (. ) 6 (. x p y x x y x y - p y (4) = p(y = 4) = p(x = 4, y = 4) = p(4,4) = (.6) 4 (.15) = .0194 p y (3) = p(3,3) + p(4,3) = 1058 . ) 15 )(. 4 (. ) 6

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## Chapter5 - CHAPTER 5 Section 5.1 1. a. b. c. P(X = 1, Y =...

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