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Ch. 5: Probability Theory Probability Assignment • Assignment by intuition – based on intuition, experience, or judgment. • Assignment by relative frequency – P(A) = Relative Frequency = f n • Assignment for equally likely outcomes Number of Outcomes Favorable to Event A P( A) Total Number of Outcomes 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 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… Outcomes • For example, to get a sum of 6, you could get: • 5, 1 4,2 3,3 … 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? Sample Space for 2 Dice 1, 1 2,1 3,1 4,1 5,1 6,1 1, 2 2,2 3,2 4,2 5,2 6,2 1, 3 2,3 3,3 4,3 5,3 6,3 1, 4 2,4 3,4 4,4 5,4 6,4 1,5 2,5 3,5 4,5 5,5 6,5 1,6 2,6 3,6 4,6 5,6 6,6 If Team A= 6, 7, 8, 9, find P(Team A) 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 Some Probability Rules and Facts • 0<= P(A) <= 1 • Think of some examples where – P(A)=0 P(A) = 1 • The sum of all possible probabilities for an experiment is 1. Ex: P(Team A)+P(Team B) =1 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 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”? Two Coins- Theoretical Answer • Outcomes: • TT, TH, HT, HH 1 2 H HH T HT H T TH TT H T 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 Three Coins • Are “1 head” , “2 heads”, and “3 heads” all equally likely? • Which are most likely and why? Three Coins 1 2 H H T T H T 2*2*2=8 outcomes 3 H T H T H T H T HHH HHT HTH HTT THH THT TTH TTT 3 coins • • • • P(0 heads)= P(1 head)= P(2 heads)= P(3 heads)= 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. Cards • 4 suits, 13 denominations; 4*13=52 cards • picture = J, Q, K A Heart (red) Diamond (red) Clubs (black) Spades (black) 2 3 4 5 6 7 8 9 10 J Q K When picking one card, find… • • • • • P(heart)= P(king)= P(picture card)= P(king or queen)= P(king or heart)= 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 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) • 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) 2 cards (independent) -questions • Example: Pick two cards, WITH replacement from a deck of cards, • P(king and king)= • P(2 hearts) = 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 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 Conditional Probability Wore seat belt No seat belt Total Driver survived 412,368 162,527 574,895 Driver died 510 1601 2111 Total 412,878 164,128 577,006 Find: P(driver died)= P(driver died/given no seat belt)= P(no seat belt)= P(no seat belt/given driver died)= Wore No seat seat belt belt Total Driver 412,368 survived 162,527 574,895 Driver died 510 1601 2111 Total 412,878 164,128 577,006 • 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 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? 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. • main – – – – Beef Beef Beef … vegetable potato dessert broc broc broc baked baked fries chocolate strawb chocolate – 4*2*2*2=32 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 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, vicepresident, and treasurer. LIST all of the different ways that this is possible. 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. AB BA CA DA AC BC CB DB AD BD CD DC 4*3=12 or 4P2 = 12 Answers 2. From these 4 people (Anne, Bob, Cindy, Dave), we wish to elect a president, vicepresident, and treasurer. LIST all of the different ways that this is possible. ABC ABD… • A B C D • B A C D • C A B A • D A B C 4*3*2 = 24 outcomes Or 4P3 = 24 C D B D A C C D A D A C B D A D B C B C A C A B ABC ABD ACB ACD BDA BDC BAC BCD BCA BCD BDA BDC CAB CAD CBA CBD DAB DAC DAB DAC DBA DBC DCA DCB 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. 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 AC BC AD BD CD 4C2= 6 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. ABC BCD ABD ACD 4C3 = 4 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 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? 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? 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. 4. 10^9 9. 5^12 5. 7! Or 7P7 10. 13 C5 20P12