Second Problem Assignment

Transcription

1 EECS 401 Due on January 19, 2007 PROBLEM 1 (30 points) Bo and Ci are the only two people who will enter the Rover Dog Fod jingle contest. Only one entry is allowed per contestant, and the judge (Rover) will delare the one winner as soon is he receivers a suitable inane entry, whih may be never. Bo writes inane jingles rapidly but poorly. He has probabilitly 0.7 of submitting his entry first. If Ci has not already won the contest, Bo s entry will be declared the winner with probability 0.3. Ci writes slowly, but he has a gift for this sort of thing. If Bo has not already won the cntest by the time of Ci s entry, Ci will be declared the winner with probability 0.6. Let B be the event that Bo enters firsts. Then B c is the event that Ci enters first. Let B w be the event that Bo wins the contest and C w be the event that Ci wins the contest. We have the following probability tree B w B C w B c w 0.4 C c w B c 0.4 C w 0.3 B w C c w 0.7 B c w 1 Due on January 19, 2007

2 (a) What is the probability that the prize never will be awarded? Thus The event that the prize is never awarded is (B B c w C c w) (B c C c w B c w). Pr (no prize awarded) Pr(B B c w C c w) + Pr(B c C c w B c w) Pr(B c w B) Pr(C c w B B c w) + Pr(B c ) Pr(C c w B c ) Pr(B c w B c C c w) (0.7) (0.7) (0.4) + (0.3) (0.4) (0.7) 0.28 (b) What is the probability that Bo will win? The event that Bo wins is given by (B B w ) (B c C c w B w ). Thus Pr(Bo wins) Pr(B B w ) + Pr(B c C c w B w ) Pr(B w B) + Pr(B c ) Pr(C c w B c ) Pr(B w B C c w) (0.7) (0.3) + (0.3) (0.4) (0.3) (c) Given that Bo wins, what is the probability that Ci s entry arrived fisrt? Pr(Ci entered first Bo wins)) Pr( Ci entered first and Bo wins) Pr(Bo wins) Pr(Bc C c w B w ) Pr(Bo wins) (0.3) (0.4) (0.3) (d) What is the probability that the first entry wins the contest? Let B be the event that Bo enters firsts. Then B c is the event that Ci enters first. Let B w be the event that Bo wins the contest and C w be the event that Ci wins the contest. We have the following probability tree Pr(first entry wins) Pr(B B w ) + Pr(B c C w ) Pr(B w B) + Pr(B c ) Pr(C w B c ) (0.7) (0.3) + (0.3) (0.6) 0.39 (e) Sppose that the probability that Bo s entry arrived first were P instead of 0.7. Can you find a value of P for which First entry wins and Second entry does not win are independent events? 2 Due on January 19, 2007

3 Let E 1 be the event that First entry wins and E 2 be the event that Second entry does not win. Note that E 1 E 2. By definition, events E 1 and E 2 are independent if and only if Pr(E 1 E 2 ) Pr(E 1 ) Pr(E 2 ). However, note that E 1 E 2 implies that Pr(E 1 E 2 ) Pr(E 1 ). Thus, events E 1 and E 2 are independent if and only if Pr(E 1 ) Pr(E 1 ) Pr(E 2 ). This is only possible if either Pr(E 1 ) 0 or Pr(E 2 ) 1. We check both of these cases as follows: Pr(E 1 ) 0 P (0.3) + (1 P) (0.6) 0 P 2 which is not possible. Pr(E c 2 ) 0 P (0.7) (0.6) + (1 P) (0.4) (0.3)) 0 P 0.4 which is not possible. Thus we have shown that Pr(E 1 ) Pr(E 1 ) Pr(E 2 ) for all P, which means there is no value of P such that the two events are independent. PROBLEM 2 (16 points) The disc containing the only copy of your term project just got corrupted, and the disc got mixed up with three other corrupted discs that were lying around. It is equally likely that any of the four discs holds the corrupted remains of your term project. Your computer expert friend offers to have a lo ok, and you know from past experience that his probability of finding your term project from any disc is 0.4 (assuming the term project is there). Given that he searches on disc 1 but cannot find your thesis, what is the probability that your thesis is on disc i for i 1, 2, 3, 4? Let D i be the event that the thesis in disc i and F be the event that your friend finds the thesis in disc 1. Now, Pr(D i ) 0.25, i 1, 2, 3, 4 and Pr(F D 1 ) 0.4 Pr(F c D 1 ) 0.6 Pr(F D 2 ) 0 Pr(F c D 2 ) 1 Pr(F D 3 ) 0 Pr(F c D 3 ) 1 Pr(F D 4 ) 0 Pr(F c D 4 ) 1 We want to find Pr(D i F c ) for i 1, 2, 3, 4. Using Bayes Rule Substituting the values, we get Pr(D i F c ) Pr(D i) Pr(F c D i ) 4 Pr(D k ) Pr(F c D k ) Pr(D 1 F c ) k1 Pr(D 2 F c ) Pr(D 3 F c ) Pr(D 4 F c ) Due on January 19, 2007

4 PROBLEM 3 (10 points) Each represents one communication link. Link failures are independent, and each link has a probabilty of 0.5 of being out of service. Towns A and B can communicate as long as they are connected in the communication network by a least one path which contains only in-service links. Determine, in an eficient manner, the probability that A and B can communicate. Consider a similar problem of two communication links in series, with failure probabilities p 1 and p 2. We can replace it by an equivalent communication link with failure probability p given by p Pr(G c 1 G c 2) Pr(G c 1) + Pr(G c 2) Pr(G c 1 G c 2) p 1 + p 2 p 1 p 2 For two communication links in parallel, we can replace it by an equivalent communication link with failure probability p given by p Pr(G c 1 G c 2) Pr(G c 1) Pr(G c 2) p 1 p 2 Similarly three communication links in parallel can be replaced by an equivalent communication link with failure probability p given by p p 1 p 2 p 3 Thus we can simplify the network as follows 4 Due on January 19, 2007

5 1/4 1/8 (1) (2) 5/8 1/8 5/16 1/8 h (3) (4) a 51/ (5) (6) Thus the probability that A and B can communicate is PROBLEM 4 (16 points) You enter a special kind of chess tournament, whereby you play one game with each of three opponents, but you get to choose the order in which you play your opponents. You win the tournament if you win two games in a row. You know your probability of a win against each of the three opponents. What is your probability of winning the tournament, assuming that you choose the optimal order of playing the opponents? Note that the underlying probability model is not explicitly given so you will have to define it. Give reasons why your model is a reasonable one. We assume the plays with different players are independent. Define W i and L i (i 1, 2, 3) as winning and losing over the ith players respectively, and let p i denote the probability of wining over the ith player, then S { } W 1 W 2 W 3, W 1 W 2 L 3, W 1 L 2 W 3, W 1 L 2 L 3, L 1 W 2 W 3, L 1 W 2 L 3, L 1 L 2 W 3, L 1 L 2 L 3 5 Due on January 19, 2007

6 The probability law is P({W 1 W 2 W 3 }) p 1 p 2 p 3, P({W 1 W 2 L 3 }) p 1 p 2 (1 p 3 ),..., We claim that the optimal order is to play the weakest player second (the order in which the other two opponents are played makes no difference). To see this, let p i be the probability of winning against the opponent played in the ith turn. Then you will win the tournament if you win against the 2nd player (prob. p 2 ) and also you win against at least one of the two other players [prob. p 1 + (1 p 1 )p 3 p 1 + p 3 p 1 p 3 ]. Thus the probability of winning the tournament is p 2 (p 1 + p 3 p 1 p 3 ), The order (1, 2; 3) is optimal if and only if the above probability is no less than the probabilities corresponding to the two alternative orders: p 2 (p 1 + p 3 p 1 p 3 ) p 1 (p 2 + p 3 p 2 p 3 ), p 2 (p 1 + p 3 p 1 p 3 ) p 3 (p 2 + p 1 p 2 p 1 ) It can be seen that the first inequality above is equivalent to p 2 p 1, while the second inequality above is equivalent to p 2 p 3. PROBLEM 5 (10 points) A coin is tossed twice. Nick claims that the event of two heads is at least as likely if we know that the first toss is a head than if we know that at least one of the tosses is a head. Is he right? Does it make a difference if the coin is fair or unfair? How can we generalize Nick s reasoning? Note that the underlying probability model is not explicitly given so you will have to define it. Give reasons why your model is a reasonable one. The sample space is S { HH, HT, TH, TT }. A reasonable model to assume is that the two tosses are independent of each other and the probability of head in the first toss is the same as the probability of head in the second toss. Let us assume that P(Head) p. Thus the probability of each event of the sample space is given by P({HH}) p 2, P({HT}) P({TH}) p(1 p) and P({TT}) (1 p) 2. Let A be the event that the outcome is two heads, i.e. A { HH }. Let B be the event that the first toss is head, i.e. B { HH, HT }. And let C be the event at least one of the tosses is head, i.e. C { HH, HT, TH }. As stated in the problem, Nick claims that P(A B) P(A C). Now we can calculate both conditional probabilities to see if the claim is true or not. and P(A B) P(A B) P(B) p 2 p 2 + p(1 p) (1) 6 Due on January 19, 2007

7 P(A C) P(A C) P(C) p 2 p 2 + 2p(1 p) (2) Now comparing the R.H.S of (1) with the R.H.S of (2), we observe that the numerators are the same, while the denominator of (1) is less than the denominator of (2). Thus, (1) greater than (2), that is, P(A B) P(A C) Generalization: The above holds for any sets A, B, C such that A B C. PROBLEM 6 (12 points) (a) We are given a sample space S and a probability law P. Let B be an event with nonzero probability. Define the real-valued function Q on the set of events in S by Q(A). P(A B). Show that Q satisfies the three axioms of a probability law and hence is a valid probability law. Asuming > 0. Q satisfies the non-negativity axiom Q(A) Pr(A B) Pr(A B) 0 Q satisfies the nomarlization axiom: Q(S) Pr(S B) Pr(S B) 1 Q satisfies the additivity axiom. Let A 1 and A 2 be two disjoint events. Q(A 1 A 2 ) Pr(A 1 A 2 B) Pr ((A 1 A 2 ) B) Pr(A 1 B) + Pr(A 2 B) (b) Using the previous part or by direct calculations, show that P(A B, C) Using part (a), assume Q(C) > 0, we have Q(A C) Q(A C) Q(C) Pr ((A 1 B) (A 2 B)) Q(A 1 ) + Q(A 2 ) P(C A, B)P(A B). (1) P(C B) Pr(A C B) Pr(C B) Pr(A C B) Pr(C B) Pr(A B, C) 7 Due on January 19, 2007

8 and Q(A C) Q(C A)Q(A) Q(C) Pr(C A, B) Pr(A B) Pr(C B) Thus, Pr(A B, C) Pr(C A, B) Pr(A B) Pr(C B) 8 Due on January 19, 2007