# ENEE 324 Quiz Solutions

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k(1 )100k(1)6For = 0.01,P_S0,100_ = (1 )100= (0.99)100= 0.3660 (2)P_S1,99_ = 100(0.01)(0.99)99= 0.3700 (3)P_S2,98_ = 4950(0.01)2(0.99)98 = 0.1849 (4)P_S3,97_ = 161, 700(0.01)3(0.99)97= 0.0610 (5)(2) The probability a packet is decoded correctly is justP [C] = P_S0,100_+ P_S1,99_+ P_S2,98_+ P_S3,97_ = 0.9819 (6)Quiz 1.10Since the chip works only if all n transistors work, the transistors in the chip are likedevices in series. The probability that a chip works is P[C] = pn.The module works if either 8 chips work or 9 chips work. Let Ck denote the event thatexactly k chips work. Since transistor failures are independent of each other, chip failuresare also independent. Thus each P[Ck] has the binomial probabilityP [C8] =_98_(P [C])8(1 P [C])98= 9p8n(1 pn), (1)P [C9] = (P [C])9= p9n. (2)The probability a memory module works isP [M] = P [C8] + P [C9] = p8n(9 8pn) (3)Quiz 1.11R=rand(1,100);X=(R0.4).*(R0.9));Y=hist(X,1:3)For a MATLAB simulation, we rst gen-erate a vector R of 100 random numbers.Second, we generate vector X as a func-tion of R to represent the 3 possible out-comes of a ip. That is, X(i)=1 if ip iwas heads, X(i)=2 if ip i was tails, andX(i)=3) is ip i landed on the edge.To see how this works, we note there are three cases: If R(i) 0 for z = 3, 4, 5, . . ..(6) If p = 0.25, the probability that the third error occurs on bit 12 isPZ (12) =_112_(0.25)3(0.75)9= 0.0645 (10)Quiz 2.4Each of these probabilities can be read off the CDF FY(y). However, we must keep inmind that when FY(y) has a discontinuity at y0, FY(y) takes the upper value FY(y+0 ).(1) P[Y < 1] = FY(1) = 09(2) P[Y 1] = FY(1) = 0.6(3) P[Y > 2] = 1 P[Y 2] = 1 FY(2) = 1 0.8 = 0.2(4) P[Y 2] = 1 P[Y < 2] = 1 FY(2) = 1 0.6 = 0.4(5) P[Y = 1] =

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