# No of choices 286968542 120526770 probability of

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No. of Choices = (2,869,685)(42) = 120,526,770 Probability of Winning = 1/120,526,770 = .0000000083 13. Initially a probability of .20 would be assigned if selection is equally likely. Data does not appear to confirm the belief of equal consumer preference. For example using the relative frequency method we would assign a probability of 5 / 100 = .05 to the design 1 outcome, .15 to design 2, .30 to design 3, .40 to design 4, and .10 to design 5. 14. a. P(E 2 ) = 1 / 4 b. P(any 2 outcomes) = 1 / 4 + 1 / 4 = 1 / 2 c. P(any 3 outcomes) = 1 / 4 + 1 / 4 + 1 / 4 = 3 / 4 15. a. S = {ace of clubs, ace of diamonds, ace of hearts, ace of spades} b. S = {2 of clubs, 3 of clubs, . . . , 10 of clubs, J of clubs, Q of clubs, K of clubs, A of clubs} c. There are 12; jack, queen, or king in each of the four suits.

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Introduction to Probability 4 - 5 d. For a: 4 / 52 = 1 / 13 = .08 For b: 13 / 52 = 1 / 4 = .25 For c: 12 / 52 = .23 16. a. (6) (6) = 36 sample points b. . 1 2 3 4 5 6 1 2 3 4 5 6 2 3 4 5 6 7 3 4 5 6 7 8 4 5 6 7 8 9 5 6 7 8 9 10 11 10 9 8 7 6 7 8 9 10 11 12 Die 1 Total for Both Die 2 c. 6 / 36 = 1 / 6 d. 10 / 36 = 5 / 18 e. No. P(odd) = 18 / 36 = P(even) = 18 / 36 or 1 / 2 for both. f. Classical. A probability of 1 / 36 is assigned to each experimental outcome. 17. a. (4, 6), (4, 7), (4 , 8) b. .05 + .10 + .15 = .30 c. (2, 8), (3, 8), (4, 8) d. .05 + .05 + .15 = .25 e. .15 18. a. P(0) = .05 b. P(4 or 5) = .20 c. P(0, 1, or 2) = .55
Chapter 4 4 - 6 19. a/b. Use the relative frequency approach to assign probabilities. For each sport activity, divide the number of male and female participants by the total number of males and females respectively. Activity Male Female Bicycle Riding .18 .16 Camping .21 .19 Exercise Walking .24 .45 Exercising with Equipment .17 .19 Swimming .22 .27 c. P(Exercise Walking) = (28.7 57.7) .35 248.5 + = d. P(Woman) = 57.7 .67 (28.7 57.7) = + P(Man) = 28.7 .33 (28.7 57.7) = + 20. a. P(N) = 56/500 = .112 b. P(T) = 43/500 = .086 c. Total in 6 states = 56 + 53 + 43 + 37 + 28 + 28 = 245 P(B) = 245/500 = .49 Almost half the Fortune 500 companies are headquartered in these states. 21. a. P(A) = P(1) + P(2) + P(3) + P(4) + P(5) = 20 50 12 50 6 50 3 50 1 50 + + + + = .40 + .24 + .12 + .06 + .02 = .84 b. P(B) = P(3) + P(4) + P(5) = .12 + .06 + .02 = .20 c. P(2) = 12 / 50 = .24 22. a. P(A) = .40, P(B) = .40, P(C) = .60 b. P(A B) = P(E 1 , E 2 , E 3 , E 4 ) = .80. Yes P(A B) = P(A) + P(B). c. A c = {E 3 , E 4 , E 5 } C c = {E 1 , E 4 } P(A c ) = .60 P(C c ) = .40 d. A B c = {E 1 , E 2 , E 5 } P(A B c ) = .60 e. P(B C) = P(E 2 , E 3 , E 4 , E 5 ) = .80 23. a. P(A) = P(E 1 ) + P(E 4 ) + P(E 6 ) = .05 + .25 + .10 = .40 P(B) = P(E 2 ) + P(E 4 ) + P(E 7 ) = .20 + .25 + .05 = .50 P(C) = P(E 2 ) + P(E 3 ) + P(E 5 ) + P(E 7 ) = .20 + .20 + .15 + .05 = .60

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Introduction to Probability 4 - 7 b. A B = {E 1 , E 2 , E 4 , E 6 , E 7 } P(A B) = P(E 1 ) + P(E 2 ) + P(E 4 ) + P(E 6 ) + P(E 7 ) = .05 + .20 + .25 + .10 + .05 = .65 c. A B = {E 4 } P(A B) = P(E 4 ) = .25 d. Yes, they are mutually exclusive. e. B c = {E 1 , E 3 , E 5 , E 6 }; P(B c ) = P(E 1 ) + P(E 3 ) + P(E 5 ) + P(E 6 ) = .05 + .20 + .15 + .10 = .50 24. Let E = experience exceeded expectations M = experience met expectations a. Percentage of respondents that said their experience exceeded expectations = 100 - (4 + 26 + 65) = 5% P(E) = .05 b. P(M E) = P(M) + P(E) = .65 + .05 = .70 25. Let Y = high one-year return M = high five-year return a. P(Y) = 15/30 = .50 P(M) = 12/30 = .40 P(Y M) = 6/30 = .20 b. P(Y M) = P(Y) + P(M) - P(Y M) = .50 + .40 - .20 = .70 c. 1 - P(Y M) = 1 - .70 = .30 26. Let Y = high one-year return M = high five-year return a. P(Y) = 9/30 = .30 P(M) = 7/30 = .23 b. P(Y M) = 5/30 = .17 c. P(Y M) = .30 + .23 - .17 = .36 P(Neither) = 1 - .36 = .64
Chapter 4 4 - 8 27.

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