1. Probability—outline: IIntroduction to Probability ASatisfactory outcomes vs. total outcomes BBasic Properties CTerminology IICombinatory Probability AThe Addition Rule – “Or” 1. The special addition rule (mutually exclusive events) 2. The general addition rule (non-mutually exclusive events) BThe Multiplication Rule – “And” 1. The special multiplication rule (for independent events) 2. The general multiplication rule (for non-independent events) CThe Complement Rule – “Not”

2. A phenomenon is random if individual outcomes are uncertain, but there is nonetheless a regular distribution of outcomes in a large number of repetitions. Examples: Tossing a coin, rolling a die, choosing a card Randomness and probability The probability of any outcome of a random phenomenon can be defined as the proportion of times the outcome would occur in a very long series of repetitions – this is known as the relative frequency definition of probability.

3. 3 Chance behavior is unpredictable in the short run but has a regular and predictable pattern in the long run.  Random: individual outcomes are uncertain but there is a regular distribution of outcomes in a large number of repetitions  Relative frequency (proportion of occurrences) of an outcome settles down to one value over the long run. That one value is then defined to be the probability of that outcome. 1 Reason to study probability: (1) It serves as the foundation of Statistics. (For expectation, Confidence interval and Hypothesis Testing, they all need a good understanding of it.) (2) It helps us to understand random mechanism. 2.Some interesting questions: (1) Suppose you tossed a fair coin 10 times and surprisingly all you’ve got were heads. If you want to make another toss, will it more likely end up with a tail? (2) With a full deck of well shuffled cards (52 cards) facing down, if you put the top card aside without turning over and looking at it, what’s the probability that the next card that you deal is a king? Examples of Randomness and probability

4. Coin toss The result of any single coin toss is random. But the result over many tosses is predictable, as long as the trials are independent (i.e., the outcome of a new coin flip is not influenced by the result of the previous flip). First series of tosses Second series The probability of heads is 0.5 = the proportion of times you get heads in many repeated trials.

5. Probability models describe mathematically the outcomes of random processes and consist of two parts: 1) S = Sample Space: This is a set, or list, of all possible outcomes of the random process. An event is a subset of the sample space. 2) A probability assigned for each possible event in the sample space S – the assignment should make sense in a model meant to describe the real world... Probability models Example: Probability Model for a Coin Toss: S = {Head, Tail} Probability of heads = 0.5 Probability of tails = 0.5

6. Sample space  Find the sample space for  1. Tossing two coins  2. Rolling two dice There are 36 possible outcomes in S, all equally likely (given fair dice). This is S: {(1,1), (1,2), (1,3), ……etc.}

7. Probability Notation An event is denoted by A, B, C… Eg: A=getting a heart from a full deck of cards. B=getting a head when flipping a coin. If A is an event, then P(A) stands for the probability that event A occurs. It is read “the probability of A”. Probability for Equally Likely Outcomes Suppose an experiment has N possible outcomes, all equally likely. Probability that event A occurs equals:

8. Use relative frequency to find Probability (sample point) Eg1: Find the following probabilities: Pr(Red) Pr(Blue) Pr(Green) Pr(Yellow) Eg2: Roll a dice labeled 1 to 6. Which of the following is LEAST LIKELY? a)an even number? b)an odd number? c)a number greater than 4? d)a number less than 4?

9. 2) Because some outcome must occur on every trial, the sum of the probabilities for all possible outcomes (the sample space) must be exactly 1. P(sample space) = 1 Coin toss: S = {Head, Tail} P(head) + P(tail) = 0.5 + 0.5 =1  P(sample space) = 1 Probability rules 1) Probabilities range from 0 (no chance of the event) to 1 (the event has to happen). For any event A, 0 ≤ P(A) ≤ 1 Probability of getting a Head = 0.5 We write this as: P(Head) = 0.5 P(neither Head nor Tail) = 0 P(getting either a Head or a Tail) = 1 Property 1: The probability of an event that cannot occur is 0. (An event that cannot occur is called an impossible event.) Property 2: The probability of an event that must occur is 1. (An event that must occur is called a certain event.)

10. Use relative frequency to find Probability http://www.hbschool.com/activity/probability_circus/

11. A deck of playing cards Four Suits: Spade, Heart, Club, Diamond; 13 Ranks: Ace, Deuce, 3,4, 5, 6, 7, 8, 9, 10, Jack, Queen, King.

12. Eg1: Event A = the king of hearts is selected P(A)=1/52 Eg2: Event B = a king is selected P(B) = 4/52=1/13

13. Eg3: Event C = a heart is selected P(C) = 13/52=1/4 Eg4: Event D = a face card is selected P(D) =12/52=3/13

14. Relationships Among Events (not E): The event that “E does not occur.” (A & B): The event that “both A and B occur.” (A or B): The event that “either A or B or both occur”  “at least one of two occur”. Venn diagram for event E:

15. Probability rules (cont. ) 3) Addition rule for disjoint events: If events A and B are disjoint, then P(A or B) = “P(A U B)” = P(A) + P(B) Note: Two events A and B are disjoint or mutually exclusive if they have no outcomes in common and can never happen together. The probability that A or B occurs is then the sum of their individual probabilities. Example: If you flip two coins, and the first flip does not affect the second flip: S = {HH, HT, TH, TT}. The probability of each of these events is 1/4, or 0.25. The probability that you obtain “only heads or only tails” is: P(HH or TT) = P(HH) + P(TT) = 0.25 + 0.25 = 0.50 Venn diagrams: A and B disjoint A and B not disjoint

16. Coin Toss Example: S = {Head, Tail} Probability of heads = 0.5 Probability of tails = 0.5 4) Multiplication rule for independent events: If events A and B are independent, then P(A and B) = P(A)*P(B) Note: Two events A and B are independent if knowing that one occurs does not change the probability that the other occurs. Two consecutive coin tosses: P(first Tail and second Tail) = P(first Tail) * P(second Tail) = 0.5 * 0.5 = 0.25 Probability rules (cont.) Venn diagram: Event A and event B. The intersection represents the event {A and B} and outcomes common to both A and B.

17. Coin Toss Example: S = {Head, Tail} Probability of heads = 0.5 Probability of tails = 0.5 Probability rules (cont.) Venn diagram: Sample space made up of an event A and its complementary A c, i.e., everything that is not A. 5) Complement rule : the probability of an event not occurring is 1 minus the probability that is does occur. P(not A)=1-P(A) The complement of any event A is the event that A does not occur, written as A c. That is: P(not A) = P(A c ) = 1 − P(A) Tail c = not Tail = Head P(Tail c ) = 1 − P(Tail) = 0.5

18. Probability Rules – Addition Rule http://www.hbschool.com/activity/probability_circus/

19. Probability Rules – Complement Rule http://www.hbschool.com/activity/probability_circus/

20. If you draw an M&M candy at random from a bag, the candy will have one of six colors. The probability of drawing each color depends on the proportions manufactured, as described here: Q1: What is the probability that an M&M chosen at random is blue? M&M candies ColorBrownRedYellowGreenOrangeBlue Probability0.30.2 0.1 ? Q2: What is probability that a random M&M is any of red, yellow, or brown? Q3: What is probability that a random M&M is neither red, yellow, nor brown? Q4: If two cadies are drawn with replacement, what is probability that the first M&M is brown, and the second M&M is yellow?

21. M&M candies ColorBrownRedYellowGreenOrangeBlue Probability0.30.2 0.1 ? If you draw an M&M candy at random from a bag, the candy will have one of six colors. The probability of drawing each color depends on the proportions manufactured, as described here: What is the probability that an M&M chosen at random is blue? What is the probability that a random M&M is any of red, yellow, or brown? S = {brown, red, yellow, green, orange, blue} P(S) = P(brown) + P(red) + P(yellow) + P(green) + P(orange) + P(blue) = 1 P(blue)= 1 – [P(brown) + P(red) + P(yellow) + P(green) + P(orange)] = 1 – [0.3 + 0.2 + 0.2 + 0.1 + 0.1] = 0.1 P(red or yellow or orange) = P(red) + P(yellow) + P(brown) = 0.2 + 0.2 + 0.3 = 0.7 What is the probability that a random M&M is neither red, yellow, nor brown?

22. Conditional probability and the independence of events The conditional probability is the probability of an event A happens given that another event B had happened. Key words: given that. Notation: Pr(B|A) Calculation of conditional probability. There are two approaches: 1. Let event A had happened and reconstruct NEW sample space S*. Then inside S*, find prob of B. 2. Based on the multiplication law of prob 22

23. Conditional probability and the independence of events EX: Pr(getting an ace given that it is a heart in a full deck of cards)=? Notation: Pr(A|B) Calculation of conditional probability: Let event A had happened and reconstruct NEW sample space S*. Then inside S*, find prob of B. 23

24. Conditional probability and the independence of events Notation: Pr(A|B) Calculation of conditional prob: two approaches: 1. Based on the multiplication law of prob 2. Let event B had happened and reconstruct NEW sample space S*. Eg1: Find prob that second toss is a head given that first toss is a head. Eg2: With a full deck of well shuffled cards, when we draw two cards out without replacement, find the prob that the 2 nd draw is a king, given that the first is a king. Eg3: suppose that a balanced die is tossed once. Find the prob of getting a 1, given that an odd number is obtained. In cases of Eg1, we claim that the two events are independent, while in Eg2, the two events are dependent! 24

25. 25 Definition/Basic Properties of Probability Example 4.1: 1. Roll a fair die twice, find out the chance that (1) the first roll is an Ace. (2) The first roll is not an Ace. (3) Both rolls are Deuces. (4) The 1st roll is an Ace and the 2nd roll is a Deuce. (5) The 2nd roll is an Ace. (6) The sum of the two rolls is 5. (7) The sum of the two rolls is at least (≥) 5. (8) The sum of the two rolls is 7.

26. Dice You toss two dice. Q: What is the probability of the outcomes summing to 5? There are 36 possible outcomes in S, all equally likely (given fair dice). Thus, the probability of any one of them is 1/36. P(the roll of two dice sums to 5) = P(1,4) + P(2,3) + P(3,2) + P(4,1) = 4 / 36 This is S: {(1,1), (1,2), (1,3), ……etc.} Q7: Prob that the sum of the two rolls is at least (≥) 5?

27. Dice: You toss two dice. Q9: What is the probability of getting a sum of 7, given that one die shows a 5? There are 11 possible outcomes in new sample space S*, all equally likely (given fair dice). Thus, the probability of any one of them is 1/11. Pr(getting a sum of 7, given that one die shows a 5) = 2/11. Two cases: (2, 5) and (5, 2). This is S*: {(1,5), (5,1), (2,5), (5,2), ……,etc.}

28. Conditional probability and independence of events http://www.youtube.com/watch?v=vhToKaPwKE4 If A and B are independent, then P(A|B)=P(A), and P(B|A)=P(B).

29. Summary to probability rules RuleKey word Key idea/ Key conditions Prob rules SumORDisjoint / Mutually exclusive Pr(A or B) = Pr(A)+Pr(B); Pr(A and B) = 0. Not disjoint / mutually exclusive Pr(A or B) = Pr(A)+Pr(B) -P(A and B) multiplicationAND (both) (ALL) Independent (with replacement) Pr(A and B) = Pr(A)٠Pr(B); Pr(B|A) = Pr(B). Dependent (without replacement) Pr(A and B) = Pr(A)٠Pr(B|A) complementNOTat least / at mostPr( not A) = 1 - Pr(A)

30. 30 Summary: Definition/Basic Properties of Probability Approach to calculate the probability of an event, say event A and denote the probability as Pr(A): 1.When the outcomes of an experiment are equally likely, then Pr(A)= 2. Pr(A)=1-Pr(not A). 3. Multiplication Rule: Pr(A and B) = Pr(A)٠Pr(B|A) For any events A and B, if A and B are independent (the outcome of A has no influence on the outcome of B, e.g Drawing tickets out with replacement is an example of independency), then Pr(A and B) = Pr(A)٠Pr(B) 4. Addition Rule: Pr (A or B) = Pr(A)+Pr(B)-P(A and B). For any events A and B, if A and B are disjoint (A and B do not share any possible common outcome, ie A and B cannot happen at the same time), then Pr (A or B) = Pr(A)+Pr(B). Note: How to find Pr(B|A)? It denote the probability of the event B given that event A is known to have occurred. The trick to find it is: Let A happen first and find out the resulting Sample Space. Then calculate the prob of B in the NEW sample space. Or use the multiplication rule: Pr(B|A)=Pr(A and B)/Pr(A).

31. 31 Definition/Basic Properties of Probability Example 2. Draw two cards out without replacement from a well shuffled full deck of cards. Find out the chance that (1) The first draw is a King. (2) The first draw is an Ace or a Jack. (3) The first draw is a either a club, a spade, or a heart. (4) The second is a club given the first is a spade. (5) The second draw is a King given the first draw is a Ace. (6) The 1 st draw is a king and the 2 nd draw is a queen.

32. 32 Definition/Basic Properties of Probability Example 2. Draw two cards out without replacement from a well shuffled full deck of cards. Find out the chance that (7) Both draws are kings. (8) The 1st is a Club and the 2nd is a Spade. (9) Both draws are spades. (10) The top five cards are all spades. (11) The first draw is a King or a Heart. (12) The second draw is a king, if we have no idea about the first card.

33. Exercise4.2: A box contain 5 balls. Each ball is either red or blue and has a number written on it. The box is made of 2 red balls and 3 blue balls, two marked 1 and three marked 0. (1) two balls are drawn with replacement at random from the box, what is the probability that the sum of the numbers is 0? (2) two balls are drawn with replacement at random from the box, what is the probability that the sum of the numbers is 1? (3) two balls are drawn with replacement at random from the box, what is the probability that the summation is NOT 1? (4) Six balls are drawn at random from the box with replacement, find the probability that you got (a) BBBBBB, and find the probability that you got (b) BRRBRB. Which case among (a) and (b) will give a bigger chance? 33 R1, R0, B1, B0, B0 Basic Properties of Probability #1

34. Summary to probability rules RuleKey word Key idea/ Key conditions Prob rules SumORDisjoint / Mutually exclusive Pr(A or B) = Pr(A)+Pr(B); Pr(A and B) = 0. Not disjoint / mutually exclusive Pr(A or B) = Pr(A)+Pr(B) -P(A and B) multiplicationAND (both) (ALL) Independent (with replacement) Pr(A and B) = Pr(A)٠Pr(B); Pr(B|A) = Pr(B). Dependent (without replacement) Pr(A and B) = Pr(A)٠Pr(B|A) complementNOTat least / at mostPr( not A) = 1 - Pr(A)

35. Exercise4.2: A box contain 5 balls. Each ball is either red or blue and has a number written on it. The box is made of 2 red balls and 3 blue balls, two marked 1 and three marked 0. (1) two balls are drawn with replacement at random from the box, what is the probability that the sum of the numbers is 0? (9/25) (2) two balls are drawn with replacement at random from the box, what is the probability that the sum of the numbers is 1? (12/25) (3) two balls are drawn with replacement at random from the box, what is the probability that the summation is NOT 1? (13/25) (4) Six balls are drawn at random from the box with replacement, find the probability that you got (a) BBBBBB, and find the probability that you got (b) BRRBRB. Which case among (a) and (b) will give a bigger chance? 35 R1, R0, B1, B0, B0 Basic Properties of Probability #2 (a)(3/5) 6 ; (b) (3/5) 3 (2/5) 3

36. Exercise4.3: A box contain 5 balls. Each ball is either red or blue and has a number written on it. The box is made of 2 red balls and 3 blue balls, two marked 1 and three marked 0. (1) two balls are drawn without replacement at random from the box, what is the probability that the second is R given the first is B? (2) one ball is drawn at random from the box, what is the probability that it is R given it is 0. (3) two balls are drawn without replacement at random from the box, what is the probability that both are red? (4) two balls are drawn without replacement at random from the box, what is the probability that the sum of the numbers is 0? 36 R1, R0, B1, B0, B0 Basic Properties of Probability #3

37. Exercise4.3: A box contain 5 balls. Each ball is either red or blue and has a number written on it. The box is made of 2 red balls and 3 blue balls, two marked 1 and three marked 0. 5) two balls are drawn without replacement at random from the box, what is the probability that the sum of the numbers is 1? 6) two balls are drawn without replacement at random from the box, what is the probability that the summation is at most (≤) 1. 7) two balls are drawn without replacement at random from the box, what is the probability that the 2 nd ball is blue without the knowledge of the 1 st ? 37 R1, R0, B1, B0, B0 Basic Properties of Probability #4

38. 38 R1, R0, B1,B0,B0 Exercise4.4 continue: A box contain 5 balls. Each ball is either red or blue and has a number written on it. The box is made of 2 red balls and 3 blue balls, two marked 1 and three marked 0. (1) two balls are drawn without replacement at random from the box, what is the probability that the second is R given the first is B? (2) one ball is drawn at random from the box, what is the probability that it is R given it is 0. (3) two balls are drawn without replacement at random from the box, what is the probability that both are blue? 6/20 (4) two balls are drawn without replacement at random from the box, what is the probability that the sum of the numbers is 0? 6/20 (5) two balls are drawn without replacement at random from the box, what is the probability that the sum of the numbers is 1? 12/20 (6) two balls are drawn without replacement at random from the box, what is the probability that the summation is at most (≤) 1. 18/20 (7) two balls are drawn without replacement at random from the box, what is the probability that the 2 nd ball is blue without the knowledge of the 1 st ? 3/5

39. 39 Exercise4.3: Mars inc. claim that the color distribution of its M&M’s is 30% for Brown, 20% each for Red and Yellow, 10% each for Blue, Orange and Green. In a package there are 40 M&M’s, (1) two M&M’s are drawn without replacement at random from the box, what is the probability that the second is Red given the first is Blue? (2) two M&M’s are drawn without replacement at random from the box, what is the probability that both are Blue? (3) Two M&M’s are drawn without replacement at random from the box, what is the probability that the 2 nd is Blue without the knowledge of the 1 st ? (4) Six M&M’s are drawn at random from the box with replacement, find the probability that you got YYYYYG, and find the probability that you got GYGGYY. ColorBrownRedYellowGreenOrangeBlue Probability0.30.2 0.1 ?

40. Exercise 4.5 : A box contain 5 balls. Each ball is either red or blue and has a number written on it. The box is made of 2 red balls and 3 blue balls. 1.two balls are drawn without replacement at random from the box, what is the probability that the second is R given that the first is B? 2.one ball is drawn at random from the box, what is the probability that it is 0 given it is B. 3.two balls are drawn without replacement at random from the box, what is the probability that both are blue? 4.two balls are drawn without replacement at random from the box, what is the probability that the sum of the numbers is 2? 5.two balls are drawn with replacement at random from the box, what is the probability that the sum of the numbers is 1? 6.two balls are drawn without replacement at random from the box, what is the probability that the 2nd ball is blue without the knowledge of the 1st? 40 R1, R0, B1,B0,B0

41. Exercise 4.6: A box contain 9 balls. Each ball is either red,green or blue and has a number written on it. The box is made of 3 red balls 4 blue balls and 2 green balls. 1.two balls are drawn without replacement at random from the box, what is the probability that the second is R given that the first is B? 3/8 2.one ball is drawn at random from the box, what is the probability that it is 0 given it is B. 1/4 3.two balls are drawn without replacement at random from the box, what is the probability that both are blue? 12/72 4.two balls are drawn without replacement at random from the box, what is the probability that the sum of the numbers is 2? 24/72 5.two balls are drawn without replacement at random from the box, what is the probability that the sum of the numbers is 0? 6/72 6.two balls are drawn with replacement at random from the box, what is the probability that the sum of the numbers is 0? 9/81 7.two balls are drawn with replacement at random from the box, what is the probability that the sum of the numbers is 1? 18/81 8.two balls are drawn without replacement at random from the box, what is the probability that the 2nd ball is blue without the knowledge of the 1st? 4/9 41 B1, B1, B2,B0, R0,R0,R2,G1,G2

42. 42 B0, B0, B1, R0, R0, R2 Exercise 4.7: A box contain 6 balls. Each ball is either red or blue and has a number written on it. The box is made of 3 red balls and 3 blue balls. 1.two balls are drawn without replacement at random from the box, what is the probability that the second is R given that the first is B? 2.one ball is drawn at random from the box, what is the probability that it is 0 given it is R. 3.two balls are drawn without replacement at random from the box, what is the probability that both are blue? 4.two balls are drawn without replacement at random from the box, what is the probability that the sum of the numbers is 2? Exercise 4.8: Let A and B be two events, 1. If Pr (A)=0.2 and Pr(B)=0.4, A and B are independent, then Pr(A & B)= _____ 2. If Pr (A)=0.2 and Pr(B)=0.4, A and B are disjoint, then Pr(A or B)= _____ (sometimes, A or B is equivalent to “at least one of two”) 3. If Pr (A&B)=0.6, Pr(A)=0.7 and Pr(B)=0.8, then Pr(A | B)= _____ Pr(B|A)=_____ HWQ: P245: 4.19, 4.20 4.25, 4.26, 4.29 4.42 P290: 4.102, 4.103,4.107,4.108,4.112,4.113,4.117