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ENGG2012B Lecture 16 Conditional probability

ENGG2012B Lecture 16 Conditional probability. Kenneth Shum. Midterm. 22 nd Mar One and a half hour. Bring calculator and blank papers. Close-book and close-note exam. Coverage: Lecture 1 to 15. Tutorial 1 to 7. Homework 1 to 3. Derangements of n=4 objects.

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ENGG2012B Lecture 16 Conditional probability

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  1. ENGG2012BLecture 16Conditional probability Kenneth Shum ENGG2012B

  2. Midterm • 22nd Mar • One and a half hour. • Bring calculator and blank papers. • Close-book and close-note exam. • Coverage: • Lecture 1 to 15. • Tutorial 1 to 7. • Homework 1 to 3. ENGG2012B

  3. Derangements of n=4 objects • Let  be the set of all 24 permutations of 1,2,3,4. • For i = 1,2,3,4, let Ai be the set of permutations which fix i. • A1 = {1234, 1243, 1324, 1342, 1423, 1432}. • A2 = {1234, 1243, 3214, 3241, 4213, 4231}. • A3 = {1234, 1432, 2134, 2431, 4132, 4231}. • A4 = {1234, 1324, 2134, 2314, 3124, 3214}. • A permutation of {1,2,3,4} not in A1 to A4 has no fixed point, and is called a derangement. ENGG2012B

  4. Venn diagram  A2 A1 2431 4231 1432 A3 4132 1234 2134 2314 1324 A4 3214 3124 1342 3241 2143 2341 2413 3142 3412 3421 4123 4312 4321 1243 4213 1423 ENGG2012B

  5. The case n=4 (cont’d) • A1  A2 = {1234, 1243}. • A1  A3 = {1234, 1432}. • A1  A4 = {1234, 1324}. • A2  A3 = {1234, 4231}. • A2  A4 = {1234, 3214}. • A3  A4 = {1234, 2134}. • A1  A2  A3 = A1  A2  A4 = A1  A3  A4 =A2  A3  A4 = A1  A2  A3  A4 ={1,2,3,4}. • By PIE, the number of derangements for n=4 is equal to 4! – 46 + 62 – 41 + 1 = 9. • Indeed, the nine derangements for n=4 are 2143, 2341, 2413, 3142, 3412, 3421, 4123, 4312, 4321. ENGG2012B

  6. Probability of no fixed number • For n=4, if we pick a permutation randomly, • the probability of no fixed number is 9/24= 0.375. • the probability of at least one fixed number is 15/24= 0.625. • For large n, it can be shown that the probability of no fixed number is approximately 1/e = 0.3679… ENGG2012B

  7. John Venn (1834-1923) • British logician and philosopher • http://en.wikipedia.org/wiki/John_Venn ENGG2012B

  8. Classical definition of probability http://en.wikipedia.org/wiki/Pierre-Simon_Laplace • From Laplace’s Théorie analytique des probabilités (1812) • “The probability of an event is the ratio of the number of cases favourable to it, to the number of all possible cases.” ENGG2012B

  9. BASIC PROBABILITY ENGG2012B

  10. Disjoint events • Two events are called disjoint if the intersection is empty, i.e., if they have no overlap. • The calculation of union of events in general is complicated when the events overlap. • It is much easier if we want to compute the probability of a union of disjoint event. We simply add the probability of the events ENGG2012B

  11. Union of disjoint events  A B C A, B, C mutually disjoint Pr(A  B  C) = Pr(A) + Pr(B) + Pr(C). ENGG2012B

  12. Partition of sample space  C A D B A  B  C  D = , A, B, C, D mutually disjoint Pr(A) + Pr(B) + Pr(C) + Pr(D)= Pr(A B  C  D) = 1 ENGG2012B

  13. Law of total probability  C A E D B A  B  C  D = , A, B, C, D mutually disjoint Pr(E) = Pr(EA)+ Pr(EB)+ Pr(EC)+ Pr(ED). ENGG2012B

  14. CONDITIONAL PROBABILITY ENGG2012B

  15. Example • There are four cards, two of them are red and two of them are black. • Shuffle the four cards and lay out the cards on the table with the front of each card facing downwards. • What is the probability that the first card is red? • We reveal the last card. • If it turns out that the last card is red, what is the probability that the first card is red? • If it turns out that the last card is black, what is the probability that the first card is red? ENGG2012B

  16. Conditional probability  • If we are given an addition information that the outcome is in event B, then we can update the likelihood to |AB|/ |B|. • If Pr(B) is nonzero, then we define the conditional probability of A given B by Pr(A|B) = Pr(AB)/Pr(B). A B At the beginning, the probability of event A is |A| / ||, assuming that all outcomes in  are equally likely. ENGG2012B

  17. The law of total probability in terms of conditional probability C A E D B  A  B  C  D = , A, B, C, D mutually disjoint Pr(E) = Pr(EA)+ Pr(EB)+ Pr(EC)+ Pr(ED) = Pr(E|A)Pr(A)+ Pr(E|B)Pr(B)+ Pr(E|C)Pr(C)+ Pr(E|D)Pr(D). ENGG2012B

  18. Example • Alice rolls a fair dice two times. • Suppose Bob is told that the face value of the first roll is 6. What is the probability the second roll is also 6? • Suppose Carol is told that the face value of one of the roll is 6. What is the probability that the other roll is also 6? ENGG2012B

  19. Independent events • If P(A|B) = P(A), then it means that the likelihood of the event A does not change after we are told that event B has occurred. In this case, event A is said to be independent of event B. • As Pr(A|B) = Pr(AB)/Pr(B), event A is independent of B if Pr(AB) = Pr(A) Pr(B). • Two events A and B are said to be independent, or statistically independent, if Pr(AB) = Pr(A) Pr(B). ENGG2012B

  20. Example • Roll two fair dice. • Let A be the event that the face value of the first die is even. • Let B be the event that the sum of the dice is odd. • Are events A and B independent? Sum is odd First numberis even ENGG2012B

  21. Pairwise independent does not imply Pr(ABC)=Pr(A)Pr(B)Pr(C). • Let  be the sample space {0,1,2,3}. The four outcomes are equally likely. • Let A be the event {0,1}, B be the event {0,2}, and C be the event {0,3}. • The three events A, B and C are pairwise independent, because • Pr(A) Pr(B) = (1/2)2 = 1/4 = Pr(AB) • Pr(B) Pr(C) = (1/2)2 = 1/4 = Pr(BC) • Pr(C) Pr(A) = (1/2)2 = 1/4 = Pr(CA) • However, • Pr(ABC) = 1/4 • Pr(A) Pr(B) Pr(C) = (1/2)3=1/8  1 A B 0 C 2 3 ENGG2012B

  22. Example • There are two boxes. • The first box contains 3 red balls and 7 blue balls. • The second box contains 16 red balls and 4 blue balls. • Perform the following random experiment • Throw a fair die. • If the number is 1, pick a ball randomly from the first box. • If the number is 2 to 5, pick a ball randomly from the second box. • What is the probability that a red ball is drawn? • What is the probability that a blue ball is drawn? • Given that a red ball is picked, what is the probability that the first box is chosen? ENGG2012B

  23. The Bayes’ rule http://en.wikipedia.org/wiki/Bayes'_theorem • Let A and B be two events. Suppose that the probability of B is nonzero. • Probability of A given B can be computed from the probability of B given A by ENGG2012B

  24. Thomas Bayes (1701-1761) • English minister and mathematician • http://en.wikipedia.org/wiki/Thomas_Bayes ENGG2012B

  25. The Monty Hall Problem • Quote from wikipedia • http://en.wikipedia.org/wiki/Monty_Hall_problem Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? • An explanation from youtube • http://www.youtube.com/watch?v=mhlc7peGlGg ENGG2012B

  26. Explanation using conditional probability • We use • “100” to represent “prize behind the first door”. • “010” to represent “prize behind the second door”. • “001” to represent “prize behind the third door”. • The host may need to flip a coin in order to open the door. We use “H” and “T” for the outcome of the coin toss. • Set up a sample space  of six elements  = {100H, 100T, 010H, 010T, 001H, 001T}. Each outcome is equally probable. ENGG2012B

  27. Before any door is opened by the host • Suppose you choose door 1 The probability thatthere is a car behind thefirst door is 1/3. 100 1/3 1/3 010 1/3 001 ENGG2012B

  28. After a door is opened • Suppose you choose door 1. • A door is opened by the host of the game. Door 3 is opened 100 H 100 T 100 Door 2 is opened 1/3 H Door 3 is opened 010 1/3 010 T Door 3 is opened 010 1/3 H Door 2 is opened 001 001 T Door 2 is opened 001 ENGG2012B

  29. Probability given that door 3 is opened • Suppose you choose door 1. • Suppose door 3 is opened • Pr(Prize is behind door 1 | door 3 is opened) = 1/3 • If you switching to door 2, you will win with probability 2/3 Door 3 is opened 100 H 100 T 100 1/3 H Door 3 is opened 010 1/3 010 T Door 3 is opened 010 1/3 H 001 001 T 001 ENGG2012B

  30. Probability given that door 2 is opened • Suppose you choose door 1. • Suppose door 3 is opened • Pr(Prize is behind door 1 | door 2 is opened) = 1/3 • If you switching to door 3, you will win with probability 2/3 100 H 100 T 100 Door 2 is opened 1/3 H 010 1/3 010 T 010 1/3 Door 2 is opened H 001 001 T Door 2 is opened 001 ENGG2012B

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