hw2_soln

hw2_soln - Probability and Stochastic Processes: A Friendly...

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Unformatted text preview: Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Roy D. Yates and David J. Goodman Problem Solutions : Yates and Goodman,1.5.2 1.5.3 1.5.5 1.6.1 1.6.4 1.6.7 1.7.4 1.8.3 1.8.4 and 1.9.1 6, Ri si . Similarly, G j ¤ £¡ ¢ (a) Since G1 s2 s3 s4 s5 s6 and all outcomes have probability 1 6, P G1 s3 and P R3 G1 1 6 so that R 3 G1 sj 1 s6 . 5 6. The event  ¡   P R 3 G1 P G1 ¤ ¡ ¢  ¡  ¤ ¦ ¦ ¦ ¦ ¡ ¢ 1 5 ¡   ¡   P R 3 G1 ¡ i ¦ §§§ ©¨¨¨¦ ¥ Problem 1.5.2 Let si denote the outcome that the roll is i. So, for 1 (b) The conditional probability that 6 is rolled given that the roll is greater than 3 is P R 6 G3 P G3 P s6 P s4 s5 s6 16 36  ¡ ¦  ¤¦¦  ¦¦ ¡   ¡  ¡  ¢ ¡   ¡   P R 6 G3 (c) The event E that the roll is even is E s2 s4 s6 and has probability 3 6. The joint probability of G3 and E is P s4 s6 13 P G3 E  The conditional probabilities of G3 given E is ¡  ¡     ¡    2 3 P E G3 P G3 2 3 P G3 E P G3 E PE 13 12 (d) The conditional probability that the roll is even given that it’s greater than 3 is Problem 1.5.3 Since the 2 of clubs is an even numbered card, C2 E so that P C2 E 2 3, P C2 E 13 12 P C2 E PE 23 P C2 1 3. Since P E  ¡  ¡    ¡  ¡     ¡     The probability that an even numbered card is picked given that the 2 is picked is P C2 E P C2 13 13 ¡  ¡     ¡    P E C2 1 1 ¡  13 12 ¡  ¡     ¡    P E G3 Problem 1.5.5 written i jk where the first card drawn is i, the second is j and the third is k. The sample space is 234 243 324 342 423 432 ¤ ¦ ¤¦¦ ¤¦¦ ¤¦¦  ¦¦¦¦ ¢ ¡ S and each of the six outcomes has probability 1 6. The events E1 E2 E3 O1 O2 O3 are O3 243 423 ¦ 234 324 342 432 234 432 ¦ O2 ¢¡ ¢¡ ¢ ¡ 243 324 342 423 324 342 ¦ ¦ ¦ ¦ ¢¡ ¢¡ ¢ ¡ E3 O1 ¦¦ E2 234 243 423 432 ¤ ¤ ¤ ¦¦¦ E1 (a) The conditional probability the second card is even given that the first card is even is P E2 E1 P E1 P 243 423 P 234 243 423 432 26 46  ¡  ¡  ¦ ¦¦  ¦  ¡     P E2 E1 12 ¡ (b) The conditional probability the first card is even given that the second card is even is P E1 E2 P E2 P 243 423 P 243 324 342 423 26 46  ¡  ¡  ¦ ¦¦  ¦  ¡     ¡    P E1 E2 12 (c) the probability the first two cards are even given the third card is even is P E1 E2 E3 P E3 ¡    ¡  P E1 E2 E3 0  (d) The conditional probabilities the second card is even given that the first card is odd is P O1 E 2 P O1 P O1 P O1 ¡   ¡     ¡    P E 2 O1 1 (e) The conditional probability the second card is odd given that the first card is odd is P O1 O2 P O1 ¡  ¡   P O2 O1 0 Problem 1.6.1 This problems asks whether A and B can be independent events yet satisfy A B? By definition, P A P B . We can see that if A B, that is events A and B are independent if and only if P AB they are the same set, then ¡   ¡  P AA PA PB §  ¡   ¡   ¡   Thus, for A and B to be the same set and also independent, PA P AB PAPB PA !  ¡     ¡   ¡    There are two ways that this requirement can be satisfied: PA 0 implying A B S. B φ. ¡¡ ¡  " ¡  " 1 implying A ¡¡ PA 2 2 ¡ P AB Problem 1.6.4 / 0, P A B 0. ¡ #  (a) Since A B  ¡# (b) To find P B , we can write PA PB PA B 58 38 PB 0 ' (  %  ¡   #  (  &  ¡  $  ' % Thus, P B PA B 1 4.  ¡ (c) Since A is a subset of Bc , P A Bc PA (d) Since A is a subset of Bc , P A Bc P Bc 3 8.  ¡  ¡ $   ¡  ¡ #  3 4. (e) The events A and B are dependent because 0  ¡    ¡  )¡ ¡   (f) Since C and D are independent P CD P CD PC PD PC D % &  ¡  $  (h) Since C and D are independent events, P C D (i) P C Dc PC PC D (j) P C Dc PC PD ¡ # ¡ #  ¡ 13 P C Dc # '&  0  ¡  $  % ' #  0  ¡  #  (k) By De Morgan’s Law, Cc 12 Dc P Cc Dc 23 12 23 13  ¡  $  ' ¡  $  $ %  ¡ ¡'  '  0  PC 13 12 56 '  ¡ 1 '   ¡ ¡  ¡  '  %  ¡ # PC D PAPB P C P D . So PD (g) This permits us to write 3 32  ¡  ¡    §    ¡  P AB PC 1 2. 16 12 1 23 16 23 C D c . This implies PC D c 1 PC D 16  (l) By Definition 1.6, events C and Dc are independent because P C Dc 16 P C P Dc 12 13     ¡    ¡  ¡  #   Problem 1.6.7 (a) For any events A and B, we can write the law of total probability in the form of P AB ' ¡     (  ¡     ¡   ¡ Since A and B are independent, P AB P ABc PA P ABc     ¡ !  '      0 % PA P A P B . This implies PAPB Thus A and Bc are independent. 3 PA 1 PB P A P Bc (b) Proving that Ac and B are independent is not really necessary. Since A and B are arbitrary labels, it is really the same claim as in part (a). That is, simply reversing the labels of A and B proves the claim. Alternatively, one can construct exactly the same proof as in part (a) with the labels A and B reversed. (c) To prove that Ac and Bc are independent, we apply the result of part (a) to the sets A and Bc . Since we know from part (a) that A and Bc are independent, part (b) says that Ac and Bc are independent. Problem 1.7.4 The tree for this experiment is 5 BH H 5 T 38 38 BT 18 3 14 AT 3 34 18 3 4 434 4 4 4 3 3 2 2 232 2 2 4 434 4 4 4 2 2 232 2 2 B T AH 5 12 34 5 A H 3 14 12 The probability that you guess correctly is PC P BH 38 38 34 %  ¡  %  ¡   (  ¡   Problem 1.8.3 P AT (a) The experiment of picking two cards and recording them in the order in which they were selected can be modeled by two sub-experiments. The first is to pick the first card and record it, the second sub-experiment is to pick the second card without replacing the first and recording it. For the first sub-experiment we can have any one of the possible 52 cards for a total of 52 possibilities. The second experiment consists of all the cards minus the one that was picked first(because we are sampling without replacement) for a total of 51 possible outcomes. So the total number of outcomes is the product of the number of outcomes for each sub-experiment. 52 51 2652 outcomes ¡6 (b) To have the same card but different suit we can make the following sub-experiments. First we need to pick one of the 52 cards. Then we need to pick one of the 3 remaining cards that are of the same type but different suit out of the remaining 51 cards. So the total number outcomes is 52 3 156 outcomes ¡6 (c) The probability that the two cards are of the same type but different suit is the number of outcomes that are of the same type but different suit divided by the total number of outcomes involved in picking two cards at random from a deck of 52 cards.  ¡ 4 156 2652 ¡ P same type, different suit 1 17 (d) Now we are not concerned with the ordering of the cards. So before, the outcomes K 8 and 8 K were distinct. Now, those two outcomes are not distinct and are only considered to be the single outcome that a King of hearts and 8 of diamonds were selected. So every pair of outcomes before collapses to a single outcome when we disregard ordering. So we can redo parts (a) and (b) above by halving the corresponding values found in parts (a) and (b). The probability however, does not change because both the numerator and the denominator have been reduced by an equal factor of 2, which does not change their ratio. 8¦7  7 ¦8 Problem 1.8.4 In this case, the designated hitter must be chosen from the 15 field players. We can break down the experiment of choosing a starting lineup into two sub-experiments. The first is to choose 1 of the 10 pitchers, the second is to choose the remaining 9 batting positions out of the 15 field players. Here we are concerned about the ordering of our selections because a new ordering specifies a new starting lineup. So the total number of starting lineups when the DH is selected among the field players is 10 1 9 is read as 15 ”permute” 9 and is equal to  15! 9! 1 816 214 400 18 162 144 000 different lineups ¦¦¦ ¡ Problem 1.9.1 9 ¡  This gives a total of N1 15 ¦¦¦¡ 9  Where 15 15 (a) Since the probability of a zero is 0.8, we can express the probability of the code word 00111 as 2 occurrences of a 0 and three occurrences of a 1. Therefore P 00111 08 2 02 3 0 00512  § ¡ §  § ¡   (b) The probability that a code word has exactly three 1’s is 5 3 5 08 2 02 3 0 0512 § ¡ § § P three 1’s ¡  ...
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This note was uploaded on 02/11/2012 for the course EEE 350 taught by Professor Duman during the Spring '08 term at ASU.

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