175_Prob and Stat for Eng Soln_Probability and Statistics for Engineering and the Sciences 6TH ED

# 175_Prob and Stat for Eng Soln_Probability and Statistics for Engineering and the Sciences 6TH ED

This preview shows pages 1–4. Sign up to view the full content.

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)=.46 175

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Chapter 5: Joint Probability Distributions and Random Samples 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 176
Chapter 5: Joint Probability Distributions and Random Samples 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 (. 3 4 ) 25 (. ) 6 (. 3 3 = + p y (2) = p(2,2) + p(3,2) + p(4,2) = ) 25 )(. 4 (. ) 6 (. 2 3 ) 3 (. ) 6 (. 2 2 + 2678 . ) 15 (. ) 4 (. ) 6 (. 2 4 2 2 = + p y (1) = p(1,1) + p(2,1) + p(3,1) + p(4,1) = ) 3 )(. 4 )(. 6 (. 1 2 ) 2 )(. 6 (. + 3590 . ) 15 (. ) 4 )(. 6 (. 1 4 ) 25 (. ) 4 )(. 6 (. 1 3 3 2 = + p y (0) = 1 – [.3590+.2678+.1058+.0194] = .2480 7. a. p(1,1) = .030 b. P(X 1 and Y 1 = p(0,0) + p(0,1) + p(1,0) + p(1,1) = .120 c. P(X = 1) = p(1,0) + p(1,1) + p(1,2) = .100; P(Y = 1) = p(0,1) + … + p(5,1) = .300 d. P(overflow) = P(X + 3Y > 5) = 1 – P(X + 3Y 5) = 1 – P[(X,Y)=(0,0) or …or (5,0) or (0,1) or (1,1) or (2,1)] = 1 - .620 = .380 e. The marginal probabilities for X (row sums from the joint probability table) are p x (0) = .05, p x (1) = .10 , p x (2) = .25, p x (3) = .30, p x (4) = .20, p x (5) = .10; those for Y (column sums) are p y (0) = .5, p y (1) = .3, p y (2) = .2. It is now easily verified that for every (x,y), p(x,y) = p x (x) p y (y), so X and Y are independent.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 30

175_Prob and Stat for Eng Soln_Probability and Statistics for Engineering and the Sciences 6TH ED

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online