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Midterm#1Sol - THE UNIVERSITY OF TEXAS AT AUSTIN Dept of...

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Unformatted text preview: THE UNIVERSITY OF TEXAS AT AUSTIN Dept. of Electrical and Computer Engineering Midterm #1 Date: October 5, 2010 Course: EE 351K Name: Last, First 0 The exam will last 75 minutes. a Closed book and notes. You may consult only one double sided page of handwritten notes. 0 No calculators are allowed. 0 All work should be performed on the exam itself. The backs of the pages may be used as scrap paper. To avoid confusion, please highlight your final solution in the space provided. 0 Fully justify your answers for more rational grading. 0 Good luck! Problem 1. [10 points] We play the following game: first, we toss a coin with P(tails) = q. If the coin comes up heads, we roll a 4-sided die; otherwise, we roll a 6-sided die. Let X be a random variable indicating the outcome of the coin toss: X = 0 if the toss is heads, X 2 1 if the toss is tails. Moreover, let Y denote the random variable corresponding to the outcome of the roll of a die (note: Y takes values from the set Sy = {1, 2, 3, 4, 5, 6}). (a) (3 points) Determine the joint probability mass function p Xy. (r i “Pm mm = § at m, ye rims} (T I (l) 7‘ GHAW'WKISQ (b) (3 points) Find E Which value of q results in E [Y] = 3? yet”) ,. A ' %+ it? , Ye\i,2,e,kg [Fl Lll : :élhv‘lm'i‘fl 5’ f7: , 7e liar}: i O’B'Lerwusc, r6 EU}; .inywtrhzer}, (%f1}3*5(%+a~3 (c) (4 points) Determine the conditional probability mass function pxw. If in some trial of this game we are told that the outcome of the die roll is Y s 2, what is the probability that the coin toss was tails? flit. ‘Krmfio 219+ 1% I l _. Q“ '3 z; $43 \fe; l‘flilfiihk TX“ 7 \ t L A . X A. Yeisléi ’FM‘K’ (Ml “1% SWAN Z c. , 3 J: P13 5 6 ii Atha’észl/ww A pm U s 2; H :0 "393:0 \k L'Kil C: 17., , , m we PM \ l l M Ewart) RL’X:H\2 'F‘L‘ffiyx'l: oquxl+I§flT=u=~§~r szww mug; '=) 39m: Vixqu 76145139 _, Z 3 12 7 Z 4. ‘5 “Z Z) 7" Problem 2. [10 points] A factory manufactures computer chips. Each chip may turn out defective with probability (1, independent of any other chips. (a) (2 points) What is the probability that a shipment of n chips will contain It defective chips? 3L3 3;)“ €35“ o? o. MMQQQX‘ mil/V: ( ,P (t: “ii (3% Mk 2 AIM {A MOXXMAQ (b) (3 points) Suppose that chips are tested in sequence, one by one, until a total of k defective chips are discovered. Find the probability that the kth defective Chip is found after exactly n chips are sampled. (Pl '2 ? h’VfiO“ KN *1 1 “t l3 livfl (Xe-Seduce 7M, $5.4; ng‘) WWW § i; ‘ A U ~01)ka 1’ JV M4 (Law Aie§ad07we engi/er E}1\CLC.E\L‘ M ; ~= i .3; (“A c-L‘k. U _énwv!¢, V~=~l (C) (5 points) Now assume that the testing procedure is unreliable, P(test says chip OK I chip defective) = P(test says chip defective | chip OK) = t, , for some 0 <2 t < 1. Suppose that each chip is tested 10 times independently, and rejected if at least 6 0f the tests come up defective. Given that a chip is rejected, find the probability that it is defective. A: iv (5 6V (wee (Loewe w? Aggfigklfwe We, sway/dc {My}; {Me Sta? L5 guea‘limw: ?(bl k ‘1 z. Wm M WNW mm? (Ew‘lfis mm) “P LM Lei llv’l; E ptolm fes‘h Slut") Ae-gggkive cl WWW“ vi. a, Agustwe WM 3 I? (Mli 't “them? (We in mm M fl'dicre'we) Re. ‘0 _ M mp“ V . , , «to-w : CL 7: (WA @423 l: Jr. (“an L (XVQMQL,M \ mag 5er to? “AM l 1 k” we“? . e M . P r wj‘wmwmww 9, M ‘k 0 MA. ~31 " 1 0 M M" cl L "fmiwa l° e u A) L 3;) i (H) M15 jwze’ Problem 3. [10 points] Consider a very simple communication network with only two nodes A and B, connected with a single link. At any given time, the link is up with probability :— 02. Starting from time t = 0, node A attempts to transmit a data packet to node B. The transmission attempts are exactly 1 second apart (1.6., the first transmission attempt is at precisely t = 1 second, the next one is at t = 2 seconds, etc.). Let Y denote the time that node B waits for the arrival of the packet. [Note: if the link is up when node A attempts to send the packet, the packet is transmitted instantaneously] (a) (3 points) What is the expected time node B waits for the arrival of the packet? E {:2 (3% gthCBS (b) (3 points) To improve the reliability of the network, another link is added in parallel to the first one. At any given time, this new link is up with probability 0 < q < 1. The packet can be transmitted if either one of the links is up. We are told that the expected time node B waits for the arrival of the packet is E[Y] = 169 seconds. Find q. i \M ?C\TV¢M WM 2 A c 0"?) U7») W 291*G’gCAHZ): (c) (4 points) Consider again the single link scenario. We are told that node B waits at least 4 seconds for the packet arrival (i.e., Y _>_ 4). Given this information, what is the expected time node B waits for the packet (i.e., find E[Y1Y 2 4])? \LD Q CEvaxk, E‘Ck‘tcwk‘b Qt "‘ winkegh‘ic RV : (Lu/mi “SA/Uta»; slam .Tgrw*g RS4“ Q I EE<V<2u1= s+vg 1% fistie Q \MQ \Lovwg V 9‘3 (LU/1.3m "e3? 7 1i>i~ge§f Miami/«Maia? (mi “i _ _ M Mir ‘Y‘x‘i’g/HLA"; r r V PP Y E “w m M35 96 a “i - i “H Epic-5 “s “N t _ M E \"< \‘t >, H 1 - M90 ti x mm?) a igfttlrsnm—fl ea “f7” Qi‘ $4 avaww +tZflUw5? X/«i Q7,“ 0 r J by?“ ' E n 1 3 £— °‘ 3;?) P AWCK'?) ...
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