# Recitation 2006

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Massachusetts Institute of TechnologyDepartment of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006)Recitation 1 February 9, 2005 1. Problem 1.2, page 52 of text. Let A and B be two sets. (a) Show the following two equalities Ac = (Ac B) (Ac B c ), B c = (A B c ) (Ac B c ) (b) Show that (A B)c = (Ac B) (Ac B c ) (A B c )

(c) Consider rolling a six-sided die. Let A be the set of outcomes where the roll is an odd number. Let B be the set of outcomes where the roll is less than 4. Calculate the sets on both sides of the equality in part (b), and verify that the equality holds. 2. Problem 1.5, page 53 of text. Out of the students in a class, 60% are geniuses, 70% love chocolate, and 40% fall into both categories. Determine the probability that a randomly selected student is neither a genius nor a chocolate lover. 3. Example 1.5, page 13 of text. Romeo and Juliet have a date at a given time, and each will arrive at the meeting place with a delay between 0 and 1 hour, with all pairs of delays being equally likely. The rst to arrive will wait for 15 minutes and will leave if the other has not yet arrived. What is the probability that they will meet?

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Massachusetts Institute of TechnologyDepartment of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006)Recitation 1 Solutions February 9, 2005 1. Problem 1.2, page 52 of text. See online solutions. 2. Problem 1.5, page 53 of text. See online solutions. 3. Example 1.5, page 13 of text. See solutions in text.

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Massachusetts Institute of TechnologyDepartment of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006)Recitation 2 February 14, 2006 1. A coin is tossed twice. Alice claims that the event of two heads is at least as likely if we know that the rst toss is a head than if we know that at least one of the tosses is a head. Is she right? Does it make a dierence if the coin is fair or unfair? How can we generalize Alices reasoning? 2. We are given three coins: one has head on both faces, the second has tails on both faces, and the third has a head on one face and a tail on the other. We choose a coin at random, toss it, and it comes up heads. What is the probability that the opposite face is tails? 3. Fischer and Spassky play a sudden death chess match. Each game ends up with either a win by Fischer, this happens with probability p, a win for Spassky, this happens with probability q, or a draw, this happens with probability 1 p q. The match continues until one of the players wins a game (and the match). (a) What is the probability that Fischer will win the last game of the match? (b) Given that the match lasted no more than 5 games, what is the probability that Fischer won in the rst game? (c) Given that the match lasted no more than 5 games, what is the probability that Fischer won the match? (d) Given that Fischer won the match, what is the probability that he won at or before the 5th game?

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Massachusetts Institute of TechnologyDepartment of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006)Recitation 2: Solutions February 14, 2006 1. Problem 1.12, page 55 of text. See online solutions. 2. Problem 1.13, page 55 of text. See online solutions. 3. (a) P(Fischer wins) = p + p(1 p q) + p(1 p q)2 + p = 1 (1 p q) =p p+q

We may also nd the solution through a simpler method: P(Fischer wins | Someone wins) = =Fischer wins

P(Fischer wins) P(Someone wins)p p+q

pq Spassky wins Fischer wins q Spassky wins

1-p

-q

pDraw

pDraw q

Fischer wins Spassky wins

(b) P(the match lasted no more than 5 games) = (p + q) + (p + q)(1 p q) + (p + q)(1 p q)2 + (p + q)(1 p q)3 + (p + q)(1 p q)4 5 = (p+q)[1(1pq) ] 1(1pq) = 1 (1 p q)5 P(Fischer wins in the rst game the match lasted no more than 5 games) = p Therefore, P(Fischer wins | the match lasted no more than 5 games) wins the match lasted no more than 5 = P(Fischer the match lasted no more than 5 games) games) P( p = 1(1pq)5

1pq

1pqDraw

pq

Fischer wins Spassky wins

pDraw q

Fischer wins Spassky wins

1pq1p-

q...

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Massachusetts Institute of TechnologyDepartment of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006)(c) P(the match lasted no more than 5 games) = 1 (1 p q)5 P(Fischer wins the match lasted no more than 5 games) = p + p(1 p q) + p(1 p q)2 + p(1 p q)3 + p(1 p q)4 5 = p[1(1pq) ] 1(1pq) =p[1(1pq)5 ] p+q

Therefore, P(Fischer wins | the match lasted no more than 5 games) wins the match lasted no more than 5 = P(Fischer the match lasted no more than 5 games) games) P( p = p+q (d) P(Fischer wins at or before the 5th game | Fischer wins) before the 5th = P(Fischer wins at or P(Fischer wins)game Fischer wins) 5 p = p[1(1pq) ] / p+q p+q = 1 (1 p q)5 This part may be solved by observing that the events {Fischer wins} and {the match lasted no more than 5 games} are independent (we know this from parts (a) and (c)): P(the match lasted no more than 5 games | Fischer wins) = P(the match lasted no more than 5 games) = 1 (1 p q)5

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Massachusetts Institute of TechnologyDepartment of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006)Recitation 3 February 16, 2006 1. A particular class has had a history of low attendance. The annoyed professor decides that she will not lecture unless at least k of the n students enrolled in the class are present. Each student will independently show up with probability pg if the weather is good, and with probability pb if the weather is bad. Given the probability of bad weather on a given day, calculate the probability that the professor will teach her class on that day. 2. Consider two coins, a blue and a red one. We choose one of the two coins at random, each being chosen with probability 1/2. Let H1 be the event that the rst toss results in heads, and H2 be the event that the second toss results in heads. The coins are biased: with the blue coin, the probability of heads in any given toss is 0.99, whereas for the red coin it is 0.01. (a) Are the events H1 and H2 (unconditionally) independent? (b) Given that the blue coin was selected, are the events H1 and H2 (conditionally) inde pendent? 3. For each one of the following statements, indicate whether it is true or false, and provide a brief explanation. (a) If P (A | B) = P (A), then P (B | Ac ) = P (B). (b) If 5 out 10 independent fair coin tosses resulted in tails, the events rst toss was tails and 10th toss was tails are independent. (c) If 10 out 10 independent fair coin tosses resulted in tails, the events rst toss was tails and 10th toss was tails are independent. (d) If the events A1 , . . . , An form a partition of the sample space, and if B, C are some other events, then P (B | C) =n i=1

P (Ai | C)P (B | Ai ).

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Massachusetts Institute of TechnologyDepartment of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006)Recitation 3 Solutions February 16, 2006 1. Problem 1.35, page 61. See online solutions. 2. Example 1.21, page 37 of text. See solutions in text. 3. (a) True If P (A | B) = P (A), then A and B are independent. And if B is independent of A, then B is also independent of Ac . This implies, by the denition of independence: P (B | Ac ) = P (B) (b) False Since there are only 5 tails out of ten, knowledge of one coin toss provides knowledge about the other coin tosses, which means the two events are not independent. In other words, the knowledge that the rst coin toss was a tails inuences the probability that the tenth coin toss is a tails. (c) True Here, all tosses are tails, so knowledge of one coin toss provides no additional knowledge about the tenth coin toss. Therefore the two events are independent. (d) False On the left hand side of the expression, since Ai s are disjoint, P (B | C) = = = P (B C) P (C)

n P (Ai )P (B C | Ai ) i=1 n i=1

P (C) P (Ai B C) P (C)

However, the right hand side of the given expression shows,n i=1

P (Ai | C)P (B | Ai ) = =

n P (Ai C) P (B Ai ) i=1 n i=1

P (C)

P (Ai )

P (Ai B C) P (C)P (Ai )

where the last line is ONLY TRUE if the events Ai C and B Ai are independent of each other. Note also for the expression to be true, i = 1 and A1 has to be the entire sample space, i.e. P (A1 ) = 1. Therefore, the given expression only holds if Ai C and B Ai are independent and i = 1.

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Massachusetts Institute of TechnologyDepartment of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006)Recitation 4 February 23, 2006 1. The birthday problem. Problem 1.45, page 66 in the text. Consider n people who are attending a party. What is the probability that each person has a distinct birthday? Assume that each person has an equal probability of being born on each day during the year, independently of everyone else, and ignore the additional complication presented by leap years (i.e., nobody is born on February 29). 2. Recall from Lecture 4 the dierent cases that arise from the problem of selecting/sampling k balls from an urn containing n numbered balls, numbered 1 through n: Sampling with replacement and ordering Sampling without replacement and ordering Sampling without replacement and without ordering Sampling with replacement and without ordering The objective of this problem is to study the fourth case. A distinct solution may be expressed in t