Ch. 5: Probability Theory

Probability Assignment Assignment by intuition – based on intuition, experience, or judgment. Assignment by relative frequency – P(A) = Relative Frequency = Assignment for equally likely 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, 14,23,3…

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, 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)

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)=0P(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 12 HHH H THT THTH TTT

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 123 HHHHH HTHHT THHTH THTT THHTHH TTHT THTTH 2*2*2=8 outcomesTTTT

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 JQK Heart (red) Diamond (red) Clubs (black) Spades (black)

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 = 8/52 P(king or heart)= 4/ /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 beltTotal Driver survived 412,368162,527574,895 Driver died 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)=

Wore seat belt No seat belt Total Driver survived 412,368162,527574,895 Driver died Total412,878164,128577,006 P(driver died)= 2111/577,006 = 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. mainvegetablepotatodessert – Beefbrocbakedchocolate – Beefbrocbakedstrawb – Beefbrocfrieschocolate – … – 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, vice- president, 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. ABBACADA ACBCCBDB ADBDCDDC 4*3=12 or 4P2 = 12

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…

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

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 ACBC ADBDCD 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. ABCBCD 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, P *26*26*10*10*10* ! or 12P *26*26* P ^9 9. 5^ ! Or 7P C5