Multiplication Rule for 2 events. Math 425 Introduction to Probability Lecture 10. Multiplication Rule for 3 events.

Transcription

1 for 2 events Math 425 Introduction to Probability Lecture 10 Kenneth Harris kaharri@umich.edu Lemma () For any events E and F (where P (F) > 0), P (E F ) P (E) P(F E) Department of Mathematics University of Michigan February 9, 2009 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 for 3 events : 3 events We can extend the to three events. Lemma For any events E, F, G (provided P (E F G) > 0) P (E F G) P (E) P(F E) P(G E F) Proof. Use the Mutiplication Rule twice, P (E F G) P (E F) P(G E F) P (E) P(F E) P(G E F) We need P (E F G) 0 to ensure the conditional probabilities exist. An urn is filled with 6 red balls, 5 blue balls, and 4 green balls. Three balls chosen at random are removed from the urn. What is the probability that the balls are of the same color? We are interested in the events (where i 1, 2, 3) R i : ith ball drawn is red, B i : ith ball drawn is blue, G i : ith ball drawn is green. C: three balls are the same color. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32

2 continued Urn: 6 red, 5 blue, and 4 green. Use the, P (R 1 R 2 R 3 ) P (R 1 ) P(R 2 R 1 ) P(R 3 R 1 R 2 ) P (B 1 B 2 B 3 ) P (B 1 ) P(B 2 B 1 ) P(B 3 B 1 B 2 ) P (G 1 G 2 G 3 ) P (G 1 ) P(G 2 G 1 ) P(G 3 G 1 G 2 ) Since these events are mutually exclusive, P (C) General multiplication Rule The Multiplication rule is the probabilistic version of the product rule for counting. Theorem (Generalized ) Let E 1, E 2,..., E n be any events such that Then P (E 1 E 2 E n ) > 0. P (E 1 E 2... E n ) P (E 1 ) P(E 2 E 1 ) P(E 3 E 1 E 2 ) (See Ross p. 71.) P(E n E 1 E 2... E n 1 ). Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 : 6 events solution In Pick-Six Lottery: A person purchases a ticket, and can choose 6 distinct numbers in the set {1, 2, 3,..., 49}. Later a Lottery Machine picks 6 distinct numbers at random in the set {1, 2, 3,..., 49}. A winning ticket is one which matches the six numbers chosen by the Machine (in any order of selection). What are the odds of winning with one ticket? Solution. We solved this before (Lecture 5) by counting using the Product Rule for counting. We conditionalize here. Let E i be the event that there are i matches. We want to compute P (E 6 ) P (E 1 E 2... E 6 ) P (E 1 ) P(E 2 E 1 ) P(E 6 E 1... E 5 ) , 983, 816 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32

3 Dependence Sometimes, changes in the conditions of an experiment change the probability of some outcomes.. An urn has 7 red balls and 5 blue balls. The balls are well mixed. A ball is drawn, its color is noted and put aside. Compare the probability that the second ball is red (R 2 ) given that the first drawn ball is red (R 1 ) versus that it is blue (R c 1 ). P(R 2 R 1 ) 6 P(R 2 R1 c 11 ) 7 11 The probability that the second ball is red is P (R 2 ) P(R 2 R 1 ) P (R 1 ) + P(R 2 R1 c ) P (Rc 1 ) Independence Sometimes, changes in the conditions of an experiment have no effect on the probability of some outcomes.. An urn has 7 red balls and 5 blue balls. The balls are well mixed. A ball is drawn, its color is noted and returned to the urn, which is again well mixed. Compare the probability that the second ball is red (R 2 ) given that the first drawn ball is red (R 1 ) versus that it is blue (R c 1 ). P(R 2 R 1 ) 7 P(R 2 R1 c 12 ) 7 12 The probability that the second ball is red is P (R 2 ) P(R 2 R 1 ) P (R 1 ) + P(R 2 R1 c ) P (Rc 1 ) Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 Independence Suppose E and F are events with P(E F) P(E F c ). By the Partition Rule P (E) P(E F )P (F) + P(E F c )P (F c ) x [P (F ) + P (F c ) ] x where x is P(E F ) or P(E F c ) So, P (E) P(E F) and P (E) P(E F c ). By the P (E F ) P(E F) P (F) P (E) P (F ). So, P (E F) P (E) P (F) Definition (Product Rule) Events E and F and independent if and only if P (E F) P (E) P (F). Equivalently, E and F are independent if and only if P(E F ) P (E) P(E F c ) Events which are not independent are said to be dependent. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32

4 Proof of Equivalence We have already shown the P(E F ) P(E F c ) P(E F ) P (E) and P (E F ) P (E) P (F ). Conversely, suppose P (E F) P (E) P (F). By the Conditioning Rule P(E F ) P (E F) P (F) By the Conditioning and Partition Rules, P(E F c ) P (E F c ) P (F c ) P (E) P (E F ) P (F c ) P (E) (1 P (F )) P (F c ) P (E) P (F) P (F) P (E). since P (E) P (E F) + P (E F c ) P (E) A standard 52 card deck is well shuffled. Are the following events independent: E: Draw a, F: Draw an ace? Solution. E and F are independent. P(E F) 1 4 P(E F c ) Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 Extended Product Rule Extended Product Rule Three dice are thrown. Are the following events independent: E 6 : Throw a six on at least one die, S 16 : The sum of the dice is 16? Definition (Extended Product Rule) The events E 1, E 2,... (possibly infinitely many events) are independent if and only if P (E i1 E i2 E in ) P (E i1 ) P (E i2 ) P (E in ) for any finite subset of indices i 1, i 2,..., i n. Solution. E and F are dependent. Challenge: verify P(S 16 E 6 ) P(S 16 E 6 ) > 0 P(S 16 E c 6 ) 0 Equivalently, the events E 1,..., E 2,... are independent if and only if P(E in E i1 E i2 E in 1 ) P (E in ), for any finite (n 2) subset of indices i 1, i 2,..., i n. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32

5 Extended Product Rule A sequence of fair coins is flipped n times, and each outcome is equiprobable. Let E i be the event that the ith flip is heads. Are the events E 1, E 2,..., E n equiprobable? Solution. They are independent: Fix any k + 1 events. Then P(E ik+1 E i1 E ik ) 2n k 1 2 n k 1 2 P (E ik+1 ) 2n 1 2 n 1 2. It is possible that two events E and F are not independent, but they become so on the assumption that a third event G occurs. Definition Events E and F are conditionally independent given G if Equivalently, P(E F G) P(E G) P(F G). P(E F G) P(E G). Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 Proof of Equivalence Suppose P(E F G) P(E G). Then, P (E F G) P(E G) Conditioning Rule for P(E F G) P (F G) P (E F G) P(E G) P (F G) P(E F G) P(E G) P(F G) divide both sides by P (G) Suppose P(E F G) P(E G) P(F G). Then, P (E F G) P (G) P (E F G) P (F G) P(E G) P(E G) P (F G) P (G) Conditioning Rule Suppose you roll a red and blue die. Consider the events L 2 : lower score is 2, H 5 : higher score is 5. D: one die is greater than 3 and one die is less than 3. Then, (a) L 2 and H 5 are not independent, (b) L 2 and H 5 are conditionally independent given D. P(E F G) P(E G) Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32

6 (a). L 2 and H 5 are not independent P (L 2 H 5 ) P (L 2 ) P (H 5 ) (b). L 2 and H 5 are conditionally independent given D P(L 2 D) 1 2 Suppose you roll a red and blue die. Consider the events R 2 : a 2 on the red die, B 2 : a 2 on the blue die, D: one die is greater than 3 and one die is less than 3. Then, (a) R 2 and B 2 are independent, (b) R 2 and B 2 are not conditionally independent given D. P(L 2 H 5 D) 1 2 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 : Baseball : Baseball (a). R 2 and B 2 are independent: P (R 2 B 2 ) 1 36 P (R 2) P (B 2 ). (b). R 2 and B 2 are not conditionally independent given D: P(R 2 D) P (R 2 D) P (D) P(R 2 B 2 D) Compare to the Problem of points, 3.4j of Ross, p. 95. The Cubs (!!) and White Sox are playing in the World Series. The Cubs win each game with probability 0.6 (independently of the games played). What is the probability that the Cubs win the Series. (The first team to win four games wins the series.) Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32

7 : Baseball : Baseball : Baseball : Baseball Method 1. (Due to Fermat) Since the first to four wins takes the Series, there are at most 7 games (4 Cub wins to 3 Sox wins). Fermat takes the sample space to be sequences of length 7: If X is an outcome in the sample space and has k W s (so, 7 k Ls), then P (X) (0.6) k (0.4) 7 k (g 1, g 2, g 3, g 4, g 5, g 6, g 7 ) where g i W, L Any sequence with 4 W s is a Cub win, otherwise it is a Sox win (there are at least 4 Ls). Not all sequences represent actual outcomes, nor are all sequences equally likely. Why does it not matter to extend a real Series play with phantom games? The phantom games do not change the probability: If the Cubs win the series in four games, they win every extension with phantom games as well. In this sample space, any sequence with at least 4 W s is a Cubs Series win. P (Cubs win) 7 n4 ( ) 7 (0.6) n (0.4) 7 n n Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 : Baseball : Baseball Method 2. (Due to Pascal) This method allows us to treat the sample space as sequences whose length is at most 7, with 4 W s or 4 Ls. Let W n,m be the event that the Cubs win the series when they have n wins and the Sox have m wins. (where n, m 4) At the start of the Series, we want to compute W 0,0. W (0, 0) can be determined from the following 3 conditions: (a) P (W 4,n ) 1, where n 3, (b) P (W n,4 ) 0 where m 3, (c) When the teams have played n + m games, and both n, m < 4, then P (W n,m ) 0.6 P (W n+1,m ) P (W n,m+1 ). The next game is either a Cubs win or Sox win. For example, P (W 3,3 ) p P (W 3,2 ) P (W 2,3 ) Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 10 February 9, / 32 The method gives the solution W 0,