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Ch. 5: Probability Theory
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Probability Assignment Assignment by intuition – based on intuition, experience, or judgment. Assignment by relative frequency – P(A) = Relative Frequency = Assignment for equally likely outcomes
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One Die Experimental Probability (Relative Frequency) – If the class rolled one die 300 times and it came up a “4” 50 times, we’d say P(4)= 50/300 – The Law of Large numbers would say that our experimental results would approximate our theoretical answer. Theoretical Probability – Sample Space (outcomes): 1, 2, 3, 4, 5, 6 – P(4) = 1/6 – P(even) = 3/6
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Two Dice Experimental Probability – “Team A” problem on the experiment: If we rolled a sum of “6, 7, 8, or 9” 122 times out of 218 attempts, P(6,7,8, or 9)= 122/218= 56% – Questions: What sums are possible? – Were all sums equally likely? – Which sums were most likely and why? – Use this to develop a theoretical probability – List some ways you could get a sum of 6…
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Outcomes For example, to get a sum of 6, you could get: 5, 14,23,3…
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Two Dice – Theoretical Probability Each die has 6 sides. How many outcomes are there for 2 sides? (Example: “1, 1”) Should we count “4,2” and “2,4” separately?
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Sample Space for 2 Dice 1, 11, 21, 31, 41,51,6 2,12,22,32,42,52,6 3,13,23,33,43,53,6 4,14,24,34,44,54,6 5,15,25,35,45,55,6 6,16,26,36,46,56,6 If Team A= 6, 7, 8, 9, find P(Team A)
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Two Dice- Team A/B P(Team A)= 20/36 P(Team B) = 1 – 20/36 = 16/36 Notice that P(Team A)+P(Team B) = 1
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Some Probability Rules and Facts 0<= P(A) <= 1 Think of some examples where – P(A)=0P(A) = 1 The sum of all possible probabilities for an experiment is 1. Ex: P(Team A)+P(Team B) =1
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One Coin Experimental – If you tossed one coin 1000 times, and 505 times came up heads, you’d say P(H)= 505/1000 – The Law of Large Numbers would say that this fraction would approach the theoretical answer as n got larger. Theoretical – Since there are only 2 equally likely outcomes, P(H)= 1/2
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Two Coins Experimental Results – P(0 heads) = – P(1 head, 1 tail)= – P(2 heads)= – Note: These all sum to 1. Questions: – Why is “1 head” more likely than “2 heads”?
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Two Coins- Theoretical Answer Outcomes: TT, TH, HT, HH 12 HHH H THT THTH TTT
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2 Coins- Theoretical answer P(0 heads) = 1/4 P(1 head, 1 tail)= 2/4 = 1/2 P(2 heads)= ¼ Note: sum of these outcomes is 1
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Three Coins Are “1 head”, “2 heads”, and “3 heads” all equally likely? Which are most likely and why?
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Three Coins 123 HHHHH HTHHT THHTH THTT THHTHH TTHT THTTH 2*2*2=8 outcomesTTTT
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3 coins P(0 heads)= P(1 head)= P(2 heads)= P(3 heads)=
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Theoretical Probabilities for 3 Coins P(0 heads)= 1/8 P(1 head)= 3/8 P(2 heads)= 3/8 P(3 heads)= 1/8 Notice: Sum is 1.
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Cards 4 suits, 13 denominations; 4*13=52 cards picture = J, Q, K A2345678910JQK Heart (red) Diamond (red) Clubs (black) Spades (black)
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When picking one card, find… P(heart)= P(king)= P(picture card)= P(king or queen)= P(king or heart)=
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Theoretical Probabilities- Cards P(heart)= 13/52 = ¼ = 0.25 P(king)= 4/52= 1/13 P(picture card)= 12/52 = 3/13 P(king or queen)= 4/52 + 4 /52 = 8/52 P(king or heart)= 4/52 + 13/52 – 1/52 = 16/52
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P(A or B) If A and B are mutually exclusive (can’t happen together, as in the king/queen example), then P(A or B)=P(A) + P(B) If A and B are NOT mutually exclusive (can happen together, as in the king/heart example), P(A or B)=P(A) + P(B) –P(A and B)
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P (A and B) For independent events: P(A and B) P(A and B) = P(A) * P(B) In General: P(A and B) = P(A) * P(B/given A)
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2 cards (independent) -questions Example: Pick two cards, WITH replacement from a deck of cards, P(king and king)= P(2 hearts) =
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P(A and B) Example-- Independent For independent events: P(A and B) P(A and B) = P(A) * P(B) Example: Pick two cards, WITH replacement from a deck of cards, P(king and king)= 4/52 * 4/52 = 16/2704 =.0059 P(2 hearts) = 13/52 * 13/52 =.0625
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P(A and B) – Dependent (without replacement) In General: P(A and B) = P(A) * P(B/given A) Example: Pick two cards, WITHOUT replacement from a deck of cards, P(king and king)= 4/52 * 3/51 = 12/2652=.0045 P(heart and heart)= 13/52 * 12/51 = 156/2652 =.059 P(king and queen) = 4/52 * 4/51 = 16/2652
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Conditional Probability Wore seat belt No seat beltTotal Driver survived 412,368162,527574,895 Driver died51016012111 Total412,878164,128577,006 Find: P(driver died)= P(driver died/given no seat belt)= P(no seat belt)= P(no seat belt/given driver died)=
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Wore seat belt No seat belt Total Driver survived 412,368162,527574,895 Driver died 51016012111 Total412,878164,128577,006 P(driver died)= 2111/577,006 =.00366 P(driver died/given no seat belt)= 1601/164,128 =.0097 P(no seat belt)= 164,128/577,006=.028 P(no seat belt/given driver died)= 1602/2111=.76
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Multiplication Problems 1. At a restaurant, you have a choice of main dish (beef, chicken, fish, vegetarian), vegetable (broccoli, corn), potato (baked, fries), and dessert (chocolate, strawberry). LIST all possible choices. 2. A teacher wishes to make all possible different answer keys to a multiple choice quiz. How many possible different answer keys could there be if there are 3 questions that each have 4 choices (A,B,C,D). LIST them all. 3. What if there were 20 multiple choice questions with 5 choices each? Explain (don’t list). 4. With 9 baseball players on a team, how many different batting orders exist?
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Answers 1. At a restaurant, you have a choice of main dish (beef, chicken, fish, vegetarian), vegetable (broccoli, corn), potato (baked, fries), and dessert (chocolate, strawberry). LIST all possible choices. mainvegetablepotatodessert – Beefbrocbakedchocolate – Beefbrocbakedstrawb – Beefbrocfrieschocolate – … – 4*2*2*2=32
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Answers 2. A teacher wishes to make all possible different answer keys to a multiple choice quiz. How many possible different answer keys could there be if there are 3 questions that each have 4 choices (A,B,C,D). LIST them all.4*4*4=64 3. What if there were 20 multiple choice questions with 5 choices each? Explain (don’t list). 5^20 4. With 9 baseball players on a team, how many different batting orders exist? 9! = 362,880
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Permutation Examples 1. If there are 4 people in the math club (Anne, Bob, Cindy, Dave), and we wish to elect a president and vice-president, LIST all of the different ways that this is possible. 2. From these 4 people (Anne, Bob, Cindy, Dave), we wish to elect a president, vice- president, and treasurer. LIST all of the different ways that this is possible.
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Answers 1. If there are 4 people in the math club (Anne, Bob, Cindy, Dave), and we wish to elect a president and vice-president, LIST all of the different ways that this is possible. ABBACADA ACBCCBDB ADBDCDDC 4*3=12 or 4P2 = 12
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Answers 2. From these 4 people (Anne, Bob, Cindy, Dave), we wish to elect a president, vice- president, and treasurer. LIST all of the different ways that this is possible. ABC ABD…
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ABCABC DABD CBACB DACD DABDA CBDC BACBAC DBCD CABCA DBCD DABDA CBDC CABCAB DCAD BACBA DCBD ABDAB CDAC DABDAB CDAC BADBA CDBC CADCA BDCB 4*3*2 = 24 outcomes Or 4P3 = 24
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Combination Examples 1. If there are 4 people in the math club (Anne, Bob, Cindy, Dave), and 2 will be selected to attend the national math conference. LIST all of the different ways that this is possible. 2. From these 4 people (Anne, Bob, Cindy, Dave), and 3 will be selected to attend the national math conference. LIST all of the different ways that this is possible.
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Combination answers 1. If there are 4 people in the math club (Anne, Bob, Cindy, Dave), and 2 will be selected to attend the national math conference. LIST all of the different ways that this is possible. AB ACBC ADBDCD 4C2= 6
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Combination answer 2. From these 4 people (Anne, Bob, Cindy, Dave), and 3 will be selected to attend the national math conference. LIST all of the different ways that this is possible. ABCBCD ABD ACD 4C3 = 4
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Permutations and Combinations Permutations – Use when ORDER matters and NO repitition – nPr = n!/(n-r)! – Example: If 10 people join a club, how many ways could we pick pres and vp? 10P2 = 90 Combinations – Use: ORDER does NOT matter and NO repitition – nCr = n!/ [(n-r)!r!] – Example: 10 people join a club. In how many ways could we pick 2? 10C2 = 45
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Multiplication, Permutation, or Combination? 1. With 14 players on a team, how many ways could we pick a batting order of 11? 2. If license plates have 3 letters and then 4 numbers, how many different license plates exist? 3. How many different four-letter radio station call letters can be formed if the first letter must be W or K? 4. A social security number contains nine digits. How many different ones can be formed? 5. If you wish to arrange your 7 favorite books on a shelf, how many different ways can this be done?
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6. If you have 10 favorite books, but only have room for 7 books on the shelf, how many ways can you arrange them? 7. You wish to arrange 12 of your favorite photographs on a mantel. How many ways can this be done? 8. You have 20 favorite photographs and wish to arrange 12 of them on a mantel. How many ways can that be done? 9. You take a multiple choice test with 12 questions (and each can be answered A B C D E). How many different ways could you answer the test? 10. If you had 13 pizza toppings, how many ways could you pick 5 of them?
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Answers 1.14P11 =175,760,000 6. 10P7 2. 26*26*26*10*10*10*10 7. 12! or 12P12 3. 2*26*26*26 8. 20P12 4. 10^9 9. 5^12 5. 7! Or 7P710. 13 C5
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