Chapter 11 – Applications of Probability
11.1
Answer Key
Descriptions of Events
Answers 1. S = {RR, RY , RG, RP , Y Y , Y G, Y P , Y R, GG, GP , GR, GY , P P , P R, P Y , P G} 2. F = {RR, Y Y , GG, P P } . H = {RR, RY , RG, RP , Y R, GR, P R}. 3.
4. F ′ is not ge깓�ng the same color twice. F ′ = {RY , RG, RP , Y R, GR, P R, Y G, Y P , GP , GY , P Y , P G}.
5. Ge깓�ng two reds is the intersec�on of the two events. F ∩H .
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Answer Key
Chapter 11 – Applications of Probability
6. (F ∪H )′ is not ge깓�ng either a red nor the same color twice. (F ∪H )′ = {Y G, Y P , GP , GY , P Y , P G}.
7. S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} 8. J = {2, 3} and K = {3, 5, 7, 9, 11} 9.
10. K ′ is ge깓�ng an even number. K ′ = {2, 4, 6, 8, 10, 12}
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Answer Key
Chapter 11 – Applications of Probability
11. This is the intersec�on of ge깓�ng an even number (K ′) and ge깓�ng a number less than 4 (J ). Ge깓�ng an even number less than 4 is K ′∩J . K ′∩J = 2
12. This is the complement of #11. This is n ot ge깓�ng an even number less than 4. Everything in the ′
rectangle that wasn't shaded in #11 will now be shaded. (J ∩K ′) = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
13. No. There is only one way to get a sum of two (ge깓�ng a 1 on both dice). There are mul�ple ways to get a 7 (for example, you could get a 1 and a 6 or a 2 and a 5, etc.). You are more likely to get a 7
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Chapter 11 – Applications of Probability
Answer Key
than a 2. 14. Unions of events are outcomes in at least one of the events and intersec�ons of events are outcomes in both events. Both unions and intersec�ons are ways of combining events. The union of two events always includes the intersec�on of the two events. 15. E ∪E ′ is the whole sample space and E ∩E ′ is the empty set.
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Answer Key
Chapter 11 – Applications of Probability
11.2
Independent Events
Answers 1. A = {H H T , H T H , T H H }. B = {H H H , T T T }. C = {H H H , H H T , H T H , T H H , H T T , T H T , T T H }. D = {T Events C and D are complements. 2. P (A) = 83 . P (B) = 41 . P (C) = 87 .P (D) = 81 . 3. P (A∩C) = 83 . P (A)P (C) = ( 83 )( 87 ) = 4. P (B∩D) =
1 8
. P (B)P (D) = ( 41 )( 81 ) =
5. P (B∩C) = 81 . P (B)P (C) = ( 41 )( 87 ) =
21 3 64 ≠ 8
. The events are not independent.
1 1 32 ≠ 8 . The events are not independent. 7 1 32 ≠ 8
. The events are not independent.
6. Answers vary. 7. Event A is all the diamonds and all the hearts. Event A has 26 outcomes. Event B is all the spades. Event B has 13 outcomes. Event C is all the fours. Event C has four outcomes. Event D is all the diamonds. Event D has 13 outcomes. 8. P (A) =
26 52
= 21 . P (B) =
13 52
= 41 . P (C) =
4 52
=
1 13 . P (D)
=
13 52
= 41 .
9. P (A∩B) = 0. P (A)P (B) = ( 21 )( 41 ) = 81 ≠0. The events are not independent. 10. P (B∩C) =
1 52 . P (B)P (C)
1 = ( 41 )( 13 )=
1 52 . The events are independent.
11. P (A∩D) = 41 .P (A)P (D) = ( 21 )( 41 ) = 81 ≠ 41 . The events are not independent. 12. The events are not independent. 13. The events are independent. 14. Disjoint events are two events that do not share any outcomes. Independent events are two events such that one event occurring does not have an effect on the probability of the second event. 15. The two events must NOT be independent.
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Chapter 11 – Applications of Probability
11.3
Answer Key
Conditional Probability
Answers 1. P (A) = 41 , P (B) = 21 , P (D) =
1 13
2. P (B) = 21 , P (A) = 1 3. The events are NOT independent. 4. P (B) =
1 13 , P (D)
1 2
=
5. The events ARE independent. 6. P (A) = 41 , P (B) = 87 , P (C) = 21 , P (D) = 7. P (B) = 71 , P (A) =
1 2
1 2
8. The events are NOT independent. 9. P (D) = 21 , P (C) =
1 2
10. The events ARE independent. 11. P (D) = 41 , P (A) =
1 2
12. The events ARE independent. 13. Answers vary. 14. Two events are independent if 1) P (A∩B) = P (A)P (B) 2) P (B) = P (A) and P (A) = P (B) . Usually either method can be used, but the condi�onal probability method might be more intui�ve. 15. Answers vary. Students might men�on that the sample space gets restricted to only the outcomes of event B.
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Answer Key
Chapter 11 – Applications of Probability
11.4
TwoWay Frequency Tables
Answers 1.
Romantic Comedies 12 8 20
Female Male Total
Action Movies
Horror Movies
Total
10 25 35
3 6 9
25 39 64
2. P (F )≈39% 3. P (R)≈31% 4. P (R) = 60% . P (F ) = 48% . These calcula�ons are different because in the first, you know that the person likes roman�c comedies and are looking for the probability that they are female, while in the second you know that the person is female and are looking for the probability that they like roman�c comedies. 5. No, because P (R)≠P (F ). 6. P (6th and bus)≈19% 7. P (6th)≈56% 8. P (bus)≈29% 9. Answers vary. 10. No, because P (6th)≠P (bus) . 11. P (disease)≈15% 12. P (positive test)≈89% 13. P (positive test∣disease)≈96% 14. P (positive test)≈11% 15. P (positive and disease or negative and no disease) =
100+560 676
=
660 676 ≈98%.
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Answer Key
Chapter 11 – Applications of Probability
11.5
Everyday Examples of Independence and Probability
Answers 1. P (D) = 0.1% , P (D ′) = 97%, P (D) = 99.5% 2‐4: Positive Test Negative Test Total
Disease 995 5 1000
No Disease 29970 969030 999000
Total 30965 96935 1000000
5. The probability that a person has the disease given that they got a posi�ve test result. 6. P (E) =
995 30965 ≈3% . This might seem surprising because it seemed like the test for the disease was
accurate. You might think that if you got a posi�ve test result, there is a large chance you have the disease. 7. The probability that a person has the disease given that they got a nega�ve test result. 8. P (E ′) =
5 96935
= 0.000052≈0.0052% .
9. The two events are not independent. Having the disease has an effect on your chance of receiving a posi�ve test result. 10. P (G) = 60%, P (S) = 80%, P (G∩S) = 55% 11. No, because P (G)P (S)≠P (G∩S) . 12. Video Games No Video Games Total 55 13. P (G) = 60 ≈92%
Snack 55 25 80
No Snack 5 15 20
Total 60 40 100
14. The probability that Ma◔� played video games given that he had a snack. 15. P (S) =
55 80 ≈69%
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Answer Key
Chapter 11 – Applications of Probability
11.6
Probability of Unions
Answers 1. The Addi�on Rule is P (A∪B) = P (A) + P (B) − P (A∩B) . It is used to find the probability of event A or event B occurring. 2. If A and B are disjoint, then P (A∪B) = P (A) + P (B) . 3. Answers vary. You must subtract the overlap between A and B so it is not counted twice. 4. 86% 5. 90% 6. 10% 7. 65% 8. P (A) = 41 ; P (D) = 9. P (A∪D) =
16 52
=
4 13
1 13 ; P (A∩D)
=
1 52
. There is a bigger chance of picking a diamond or a four than a diamond and a
four. 10. Events B and D are disjoint, so P (B∪D) = P (B) + P (D). 11. P (C∪D) =
3 4
. This is the probability that the first coin is a heads, the third coil is a tails, or both.
12.
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Chapter 11 – Applications of Probability 13. Answers vary. P (B∪C) = 1 , P (B∩C) =
3 8
Answer Key
.
14. A∩B∩C 15. P (A∪B∪C) = P (A) + P (B) + P (C) − P (A∩B) − P (A∩C) − P (B∩C) + P (A∩B∩C)
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Chapter 11 – Applications of Probability
11.7
Answer Key
Probability of Intersections
Answers 1. The Mul�plica�on Rule is used to find the probability of intersec�ons of events. P (A∩B) = P (B)P (B) = P (A)P (A). 2. 0, because they do not share any outcomes. 3. P (A)P (B) 4. If the events are independent, then P (A∩B) = P (B)P (B) is equivalent to P (A)P (B). 5. 0.08=8% 6. ≈0.43 = 43% 7. ≈0.23 = 23% 8. 99.95% 9. ≈96.95% 10. 1% of the people with the disease will receive a nega�ve test result. P (Disease ∩N egative test) = 0.0005% . 11. 5 people out of the 1 million. 12. 27% 13. P (A∩B∩C) = P (C∩[A∩B]) = P (C∣A∩B)P (A∩B) 14. P (A∩B∩C) = P (A∩B)P (A∩B) = P (A∩B)P (A)P (A) because P (A∩B) = P (A)P (A). 15. 13.05%
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Chapter 11 – Applications of Probability
11.8
Answer Key
Permutations and Combinations
Answers 1. 8P 2 = 56 . This is the number of ways to choose and arrange 2 objects from a set of 8 objects. 2. 8P 8 = 40320 . This is the number of ways to arrange 8 objects. 3. 8C8 = 1 . This is the number of ways to choose 8 objects from a set of 8 objects. 4. 14C8 = 3003. This is the number of ways to choose 8 objects from a set of 14 objects. 5. Yes, because nCk =
nP k k!
, and k ! is always greater than or equal to 1.
6. Use your calculator to verify that 10C5 = 252. 7. Permuta�on, because order ma◔�ers. There is only one way to choose 0, 1, 2. The probability is 1 1 41P 3 = 63960 . 8. Combina�on, because order doesn't ma◔�er. There is only one way to choose Bobby and David. The 1 1 probability is 5C2 = 10 . 9. Combina�on, because order doesn't appear to ma◔�er. 12C4 = 495. 10. Permuta�on, because order ma◔�ers. There are 4P 3 = 24 ways the winners could be all boys. There 24 are 14P 3 = 2184 possible 1st, 2nd, 3rd place winners. The probability is 2184 ≈1.1%. 720 11. There are 10P 3 = 720 ways the winners could be all girls. The probability is 2184 ≈33%.
12. Combina�on, because order doesn't ma◔�er. There are 5C4 = 5 ways to choose 4 colleges within 5 100 miles. There are 12C4 = 495 ways to choose 4 colleges. The probability is 495 ≈1% . 13. Permuta�on, because order ma◔�ers. 12P 5 = 95040 is the number of ways you could rank your top 1 five. If Jesse chooses one ranking at random, he has a 95040 chance of choosing the right one. 14. Combina�on, because order doesn't ma◔�er. 10C3 = 120. Your probability of ge깓�ng any par�cular 1 3 item combo is 120 . 15. There are 4 royal flushes, 1 per suit. For the number of poker hands it is a combina�on calcula�on because order of cards within the hand doesn't ma◔�er. There are 52C5 = 2598960 poker hands. 4 The probability is 2598960 = 0.00015%
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Chapter 11 – Applications of Probability
11.9
Answer Key
Probability to Analyze Fairness and Decisions
Answers 1. A game is fair if everyone has an equal chance of winning. 2. Flipping a coin can help to make a random choice when you have two op�ons. 3. You could put names in a hat and randomly draw a name, or use a random number generator. 4. Answers vary. You could flip two coins and let the four outcomes of HH, HT, TH, TT each belong to a different person. 5. Answers vary. 6. When you want to pick a certain number of people randomly from a set of people. 7. This game is fair. 15 8. This game is not fair. The probability of you winning is 36 and the probability of your friend winning 21 is 36 .
9. The game is not fair. Your probability of winning is 92 . Shelly's probability of winning is 91 . Lisa's probability of winning is 96 . 10. In each case you have a 83 chance of winning so it doesn't ma◔�er which game you play. 19 11. Yes, because you have a greater than 50% chance of winning. Your chance of winning is 27 .
12. This game is fair. 28 24 13. The game is not fair. Rachel's chance of winning is 52 and your chance of winning is 52 . 6 8 14. The game is not fair. Your chance of winning is 24 and your friend's chance of winning is 24 . Your
friend has a greater chance of winning. 15. Yes, he has a 3/4 chance of winning if his strategy is to switch.
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