chp 10 solutions

7/27/2019 Chp 10 Solutions

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Chapter 10 Solutions10.1. In the long run, of a large number of Texas Holdem games in which you hold a pair,

the fraction in which you can make four of a kind will be about 88/1000 (or 11/125) . Itdoes not mean that exactly 88 out of 1000 such hands would yield four of a kind; thatwould mean, for example, that if youve been dealt 999 such hands and only had four of akind 87 times, then you could count on getting four of a kind the next time you held a pair.

102. (a) An impossible event ha s probability 0. (b) A certain event ha s probability 1. (c) Anevent with probability 0.01 would be called very unlikely. (d) This event ha s probability0.6 (or 0.99, but more often than not is a rather weak description of an event withprobability 0.99).

10.3. (a) There are 4 zeros among the first 50, for a proportion of 0.08. (b) Answers will vary,but more than 99% of all students should get between 7 and 33 heads out of 200 flips.

0.18-0.16-0.14-

~ 0.12- 0.08-& 0.06-

0.04-0.02-

0-

10.5. (a) S (male, female). (b) S = (All numbers between and inches). (Choicesof upper and lower limits will vary.) (c) S {all numbers greater than or equal to 0), orS (0,0.01,0.02,0.03, . . .). (d) S = (A, B , C, D, F) (students might also include + and

10.6. (a) The tab le on the right illustrates the 16 possi- _______ble pair combinations in the sample space. (b) Eachof the 16 outcomes ha s probability 1/16.

10.7. For the sample space, add I to each pair-total in the table shown inthe previous solution: S = (3,4,5,6,7,8,9). As all faces are equallylikely and the dice are independent, each of the 16 possible pairings isequally likely, so (for example) the probability of a total of 5 is 3/16,because 3 pairings add to 4 (and then we add 1). The complete se t ofprobabilities is shown in the table on the right.

10.4. (a) In almost 99% of all simulations,there will be between 5 and 15 heads, sothe sample proportion will be between0.25 and 0.75. (b) Shown on the right isthe theoretical histogram; a stemplot of 25proportions will have roughly that shape.

11111111111111~ 11110 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9Proportion of heads

GD GD0DElD

00 GD E1~1DO DO DO GD00 ND DO GDDO DO GD GDTotal Probability

3 1/164 2/165 3/166 4/167 3/168 2/169 1/16

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Solutions 145

10.8. (a) 35% (20% + 15%) are currently undergraduates. This makes use of Rule 3,because (assuming there are no double majors) undergraduate students in businessand undergraduate students in other fields have no students in common. (b) 80%(100% 20%) are not undergraduate business students. This makes use of Rule 4.

10.9. (a) Event B specifically rules out obese subjects, so there is no overlap withevent A. (b) A or B is the event The person chosen is overweight or obese:P(Aor B) = P(A) + P(B) = 0.32 + 0.34 = 0.66. (c) P(C) = I P(A orB) = 0.34.

10.10. (a) The g iven probabilities have sum 0.91, so P(other language) = 0.09. (b) P(notEnglish) = 1 0.63 = 0.37. (Or, add the other three probabilities.) (c) P(neither English norFrench) = 0.06 + 0.09 = 0.15. (Or, subtract 0.63 + 0.22 from 1. )

10.11. Model I: Not legitimate (probabilities have sum ~). Model 2: Legitimate. Model 3: Notlegitimate (probabilities have sum ~). Model 4: Not legitimate (probabilities cannot be morethan 1).

10.12. (a) A = (7,8,9), 50 P(A) 0.058 + 0.051 + 0.046 = 0.155. (b) B = (1,3,5,7,9), soP(B) = 0.301 +0.125+0.079+0.058+0.046 = 0.609. (c) AorB = (1,3,5,7,8,9), soP(A or B) = 0.301 + 0.125 + 0.079 + 0.058 + 0.051 + 0.046 = 0.660. This is different fromP(A) + P(B) because A and B are not disjoint.

10.13. (a) The eight probabilities have sum I (and account fo r all possible responsesto the question). (b) X < 7 means the subject worked out fewer than 7 daysin the past week: or the subject d id not work out every day in the past week:P(X < 7) = I 0.02 = 0.98 (or, add the first 7 probabilities). (c) Worked out at leastonce is X > 07 P(X > 0) = 1 0.68 = 0.32 (or, add the last 7 probabilities).

10.14. (a) P(Y


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14 6 Chapter 10 Introducing Probability

10.18. (a) Y ? 8 means the student runs the mile in 8 minutes or more. P(Y ? 8) =p(Y7.I l > 8~I) = P(Z ? 1.2) 0.1151 (using Table A). (b) The student could run amile in less than 6 minutes is the event Y 2.56, for which Table A gives 0.0052.1031. (a ) There are sixteen possible outcomes:

HIIHN, HHHI~1, HHMH, MMHH, MHHH, HHMM, HMHM, HMMH,MRHH, MHMH, MMHH, 1114MM, 1411MM, MMHM, 1414)411, 14)4MM)

(b) The sample space is (0,1,2,3,4).1032. (a) Legitimate. (5) Legitimate (even if the deck of cards is not!). (c) Not legitimate (the

total is more than 1).


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148 Chapter 10 Introducing Probability10.43. (a) P(2l years old or older) = 0.56the sum of the third and fourth columns.

(b) P(does not live with his/her parents) = 0.54the sum of the second and third rows (orone minus the sum of the first row).

10.44. (a) X is discrete, because it has a finite sample space. (b) At least one nonworderror is the event {X ? 1) (or {X > 0)). P(X ? 1) = I P(X = 0) = 0.9.(c) {X < 2) is no more than two nonword errors: or fewer than three nonworderrors. P(X ~ 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1 + 0.2 + 0.3 = 0.6.P(X 0.72-0.56) P(Z>8.42); this is basically 0.


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Solutions 14 910.52. P(8.9

chp 10 solutions

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