Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Roy D. Yates and David J.

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1 Probability and Stochastic Processes: A Friendly Introduction for Electrical and oputer Engineers Roy D. Yates and David J. Goodan Proble Solutions : Yates and Goodan, and Proble 1..3 The saple space is S A K A K A K A K The event H is the set H A K Proble The saple space of the experient is S LF BF LW BW Fro the proble stateent, we know that P LF 0, P BF 0 and P BW 0. This iplies P LW The questions can be answered using Theore 1.. (a) The probability that a progra is slow is PW P LW P BW (b) The probability that a progra is big is P B P BF P BW (c) The probability that a progra is slow or big is PW B PW P B P BW Proble It is tepting to use the following proof: Since S and φ are utually exclusive, and since S S φ, Since P S 1, we ust have P φ 0. 1 P S φ P S P φ 1

2 The above proof used the property that for utually exclusive sets A 1 and A, P A 1 A P A 1 P A The proble is that this property is a consequence of the three axios, and thus ust be proven. For a proof that uses just the three axios, let A 1 be an arbitrary set and for n 3, let A n φ. Since A 1 A i, we can use Axio 3 to write P A 1 P A i P A 1 P A By subtracting P A 1 fro both sides, the fact that A P φ P A i 0 n 3 P A i i 3 φ perits us to write By Axio 1, P A i 0 for all i. Thus, n 3 P A i 0. This iplies P φ 0. Since Axio 1 requires P φ 0, we ust have P φ 0. Proble Following the hint, we define the set of events A i such that, A i B i and for i, A i φ. By construction, B i A i. Axio 3 then iplies P B i P A i P A i For i, P A i 0, yielding P B i P A i P B i Proble Theore 1.7 requires a proof fro which we can check which axios are used. However, the proble is soewhat hard because there ay still be a sipler proof that uses fewer axios. Still, the proof of each part will need Theore 1.4 which we now prove. For the utually exclusive events B 1 B, let A i B i for and let A i φ for i. In that case, by Axio 3, P B 1 B B P A 1 A! 1 P A i i Now, we use Axio 3 again on A A " 1 to write! 1 P B i P A i P A i i P A i P A A " 1 P B i

3 Thus, we have used just Axio 3 to prove Theore 1.4: P B 1 B # B P B i % ' % (a) To show P φ 0, let B 1 S and let B φ. Thus by Theore 1.4 P S P B 1 B P B 1 P B P S P φ Thus, P φ 0. Note that this proof uses only Theore 1.4 which uses only Axio 3. (b) Using Theore 1.4 with B 1 A and B A c, we have P S P A A c P A$ P A c Since, Axio says P S 1, P A c 1 P A. This proof uses Axios and 3. (c) By Theore 1., we can write both A and B as unions of disjoint events: A &% AB'( AB c B &% AB' A c B' Now we apply Theore 1.4 to write P A P AB P AB c P B P AB P A c B We can rewrite these facts as P AB c P A$ P AB P A c B P B$ P AB (1) Note that so far we have used only Axio 3. Finally, we observe that A the union of utually exclusive events B can be written as Once again, using Theore 1.4, we have A B &% AB' % AB c ' % A c B' P A B P AB P AB c P A c B () Substituting the results of Equation 1 into Equation yields P A B P AB$ P A P AB P B P AB which copletes the proof. Note that this clai required only Axio 3. (d) Observe that since A ) B, we can write B as the disjoint union B A % A c B'. By Theore 1.4 (which uses Axio 3), P B P A$ P A c B By Axio 1, P A c B 0, hich iplies P A P B. This proof uses Axios 1 and 3. 3

4 Proble 1..6 The proble stateent yields the obvious facts that P L 0 16 and P H The words 10% of the ticks that had either Lye disease or HGE carried both diseases can be written as (a) Since LH ) L H, P LH L H 0 10 P LH * % L H'+ P LH L H P L H P LH P L H 0 10 Thus, Since P L 0 16 and P H 0 10, P LH 0 10P L H 0 10 % P L P H P LH,' P LH 0 10 % ' (b) The conditional probability that a tick has Lye disease given that it has HGE is P L H P LH P H 0 36 Proble (a) For any events A and B, we can write the law of total probability in the for of P A P AB P AB c Since A and B are independent, P AB P A P B. This iplies Thus A and B c are independent. P AB c P A P A P B P A % 1 P B-' P A P B c (b) Proving that A c and B are independent is not really necessary. Since A and B are arbitrary labels, it is really the sae clai as in part (a). That is, siply reversing the labels of A and B proves the clai. Alternatively, one can construct exactly the sae proof as in part (a) with the labels A and B reversed. (c) To prove that A c and B c are independent, we apply the result of part (a) to the sets A and B c. Since we know fro part (a) that A and B c are independent, part (b) says that A c and B c are independent. 4

5 > > > > Proble In the Venn diagra at right, assue the saple space has area 1 corresponding to probability 1. As drawn, A, B, and each have area 1. 3 and thus probability The three way intersection AB has zero probability, iplying A, B, and are not utually independent since ;: P AB 0 / P A P B P However, AB, B, and A each has area As a result, each pair of events is independent since P AB P A P B P B P B P P A P A P Proble 1.7. The P H is the probability that a person who has HIV tests negative for the disease. This is referred to as a false-negative result. The case where a person who does not have HIV but tests positive for the disease, is called a false-positive result and has probability P H c. Since the test is correct 99% of the tie, P H P H c 0 01 Now the probability that a person who has tested positive for HIV actually has the disease is P H < P H P = P H P H$ P H c We can use Bayes forula to evaluate these joint probabilities. P H < P H P H P H P H P H c P H c % 0 99' % 0 000' % 0 99' % 0 000'( % 0 01' % ' Thus, even though the test is correct 99% of the tie, the probability that a rando person who tests positive actually has HIV is less than 0.0. The reason this probability is so low is that the a priori probability that a person has HIV is very sall. Proble The tree for this experient is 1? > > A 1 1? B 1 A A A A1? 4A A H 1 T 3? 4 1 B 3? 4 B B B B B 1? 4 H 1 T 1 A A A A 3? 4A A H T 1? B 3? 4 B B H B B B 1? 4 T A A A A 1? 4A A H T 3? B 1? 4 B B H B B B 3? 4 T A 1 H 1 H 3? 3 A 1 H 1 T 1? 3 A 1 T 1 H 9? 3 A 1 T 1 T 3? 3 B 1 H 1 H 3? 3 B 1 H 1 T 9? 3 B 1 T 1 H 1? 3 B 1 T 1 T 3? 3

6 The event H 1 H that heads occurs on both flips has probability The probability of H 1 is Siilarly, P H 1 H P A 1 H 1 H $ P B 1 H 1 H 6. 3 P H 1 P A 1 H 1 H P A 1 H 1 T P B 1 H 1 H P B 1 H 1 T 1. P H P A 1 H 1 H P A 1 T 1 H P B 1 H 1 H P B 1 T 1 H 1. Thus P H 1 H =/ P H 1 P H, iplying H 1 and H are not independent. This result should not be surprising since if the first flip is heads, it is likely that coin B was picked first. In this case, the second flip is less likely to be heads since it becoes ore likely that the second coin flipped was coin A. Proble What our design ust specify is the nuber of boxes on the ticket, and the nuber of specially arked boxes. Suppose each ticket has n boxes and k specially arked boxes. Note that when k 0, a winning ticket will still have k unscratched boxes with the special ark. A ticket is a winner if each tie a box is scratched off, the box has the special ark. Assuing the boxes are scratched off randoly, the first box scratched off has the ark with probability % k'. n since there are k arked boxes out of n boxes. Moreover, if the first scratched box has the ark, then there are 4 k arked boxes out of n 1 reaining boxes. ontinuing this arguent, the probability that a ticket is a winner is k 4 k 3 k k 1 k % k '! % n '! p n n 1 n n 3 n 4 k!n! By careful choice of n and k, we can choose p close to For exaple, soe possible choices are n k p Probably, a gaecard with N 14 boxes and k 7 shaded boxes would be quite reasonable. Proble (a) There are 3 group 1 kickers and 6 group kickers. Using G i to denote that a group i kicker was chosen, we have P G P G. 3 In addition, the proble stateent tells us that P K G 1 1. P K G 1. 3 obining these facts using the Law of Total Probability yields P K P K G 1 P G 1 P K G P G &% 1. ' % 1. 3'( % 1. 3' %. 3'

7 (b) To solve this part, we need to identify the groups fro which the first and second kicker were chosen. Let c i indicate whether a kicker was chosen fro group i and let i j indicate that the first kicker was chosen fro group i and he second kicker fro group j. The experient to choose the kickers is described by the saple tree: A A A A 3? 9A A c 1 B B B B B B 6? 9 c A A A A? 8 A A c 1 c 6? 8 B 3? 8 B B c B 1 B B? 8 c 11 1? 1 1 1? 4 1 1? 4? 1 Since a kicker fro group 1 akes a kick with probability 1. while a kicker fro group akes a kick with probability 1. 3, P K 1 K 11 &% 1. ' P K 1 K 1 D% 1. ' % 1. 3' P K 1 K 1 &% 1. 3' % 1. ' P K 1 K D% 1. 3' By the law of total probability, P K 1 K P K 1 K 11 P 11 P K 1 K 1 P 1 P K 1 K 1 P 1 P K 1 K P It should be apparent that P K 1 P K fro part (a). Syetry should also ake it clear that P K 1 P K since for any ordering of two kickers, the reverse ordering is equally likely. If this is not clear, we derive this result by calculating P K i j and using the law of total probability to calculate P K. By the law of total probability, P K P K P K 1 1. P K 1. 3 P K P K 11 P 11 P K 1 P 1 P K 1 P 1 P K P We observe that K 1 and K are not independent since P K 1 K 1 96 / P K 1 P K

8 Note that and % 7. 18' are close but not exactly the sae. The reason K 1 and K are dependent is that if the first kicker is successful, then it is ore likely that kicker is fro group 1. This akes it ore likely that the second kicker is fro group and is thus ore likely to iss. (c) Once a kicker is chosen, each of the 10 field goals is an independent trial. If the kicker is fro group 1, then the success probability is 1.. If the kicker is fro group, the success probability is Out of 10 kicks, there are isses iff there are successful kicks. Given the type of kicker chosen, the probability of isses is P M G 1 10 % 1. ' % 1. ' P M G 10 % 1. 3' %. 3' We use the Law of Total Probability to find P M P M G 1 P G 1 P M G P G 10 % 1. 3' % 1. ' 10 %. 3' % 1. 3' %. 3' 8