3 - CHAPTER 2 Solution to Problem 2.1. Let X be the number...

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CHAPTER 2 Solution to Problem 2.1. Let X be the number of points the MIT team earns over the weekend. We have P ( X =0)=0 . 6 · 0 . 3=0 . 18 , P ( X =1)=0 . 4 · 0 . 5 · 0 . 3+0 . 6 · 0 . 5 · 0 . 7=0 . 27 , P ( X =2)=0 . 4 · 0 . 5 · 0 . . 6 · 0 . 5 · 0 . 7+0 . 4 · 0 . 5 · 0 . 7 · 0 . 5=0 . 34 , P ( X =3)=0 . 4 · 0 . 5 · 0 . 7 · 0 . 5+0 . 4 · 0 . 5 · 0 . 7 · 0 . . 14 , P ( X =4)=0 . 4 · 0 . 5 · 0 . 7 · 0 . . 07 , P ( X> 4) = 0 . Solution to Problem 2.2. The number of guests that have the same birthday as you is binomial with p =1 / 365 and n = 499. Thus the probability that exactly one other guest has the same birthday is ± 499 1 ² 1 365 ³ 364 365 ´ 498 0 . 3486 . Let λ = np = 499 / 365 1 . 367. The Poisson approximation is e λ λ = e 1 . 367 · 1 . 367 0 . 3483, which closely agrees with the correct probability based on the binomial. Solution to Problem 2.3. (a) Let L be the duration of the match. If Fischer wins a match consisting of L games, then L 1 draws must ±rst occur before he wins. Summing over all possible lengths, we obtain P (Fischer wins) = 10 X l =1 (0 . 3) l 1 (0 . 4) = 0 . 571425 . (b) The match has length L with L< 10, if and only if ( L 1) draws occur, followed by a win by either player. The match has length L = 10 if and only if 9 draws occur. The probability of a win by either player is 0.7. Thus p L ( l )= P ( L = l ( (0 . 3) l 1 (0 . 7) ,l ,..., 9, (0 . 3) 9 = 10, 0 , otherwise. Solution to Problem 2.4. (a) Let X bethenumbero fmodemsinuse .For k< 50, the probability that X = k is the same as the probability that k out of 1000 customers need a connection: p X ( k ± 1000 k ² (0 . 01) k (0 . 99) 1000 k ,k =0 , 1 49 . 21
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The probability that X = 50, is the same as the probability that 50 or more out of 1000 customers need a connection: p X (50) = 1000 X k =50 ± 1000 k ² (0 . 01) k (0 . 99) 1000 k . (b) By approximating the binomial with a Poisson with parameter λ = 1000 · 0 . 01 = 10, we have p X ( k )= e 10 10 k k ! ,k =0 , 1 ,..., 49 , p X (50) = 1000 X k =50 e 10 10 k k ! . (c) Let A be the event that there are more customers needing a connection than there are modems. Then, P ( A 1000 X k =51 ± 1000 k ² (0 . 01) k (0 . 99) 1000 k . With the Poisson approximation, P ( A )isest imatedby 1000 X k =51 e 10 10 k k ! . Solution to Problem 2.5. (a) Let X be the number of packets stored at the end of the Frst slot. ±or k<b , the probability that X = k is the same as the probability that k packets are generated by the source: p X ( k e λ λ k k ! , 1 ,...,b 1 , while p X ( b X k = b e λ λ k k ! =1 b 1 X k =0 e λ λ k k ! . Let Y be the number of number of packets stored at the end of the second slot. Since min { X, c } is the number of packets transmitted in the second slot, we have Y = X min { X, c } .Thu s , p Y (0) = c X k =0 p X ( k c X k =0 e λ λ k k ! , p Y ( k p X ( k + c e λ λ k + c ( k + c )! c 1 , 22
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p Y ( b c )= p X ( b )=1 b 1 X k =0 e λ λ k k ! . (b) The probability that some packets get discarded during the frst slot is the same as the probability that more than b packets are generated by the source, so it is equal to X k = b +1 e λ λ k k !
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3 - CHAPTER 2 Solution to Problem 2.1. Let X be the number...

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