University of Illinois Fall 2008
ECE 313: Solutions to Problem Set 6
1. Let X= bit stored, Y= bit read
(a) P(Y= 1) = P(Y= 1|X= 1)P(X= 1) + P(Y= 1|X= 0)P(X= 0)
= (0.85)(0.5) + (0.1)(0.5) = 0.475.
(b) P(X= 1|Y= 1) = P(Y=1,X=1)
P(Y=1) =P(X=1)P(Y=1|X=1)
P(Y=1) =(0.5)(0.85)
0.475 = 0.8947.
2. (a) P(T+) = P(T+|H+)P(H+) + P(T+|H−)P(H−).
Test A: P(T+) = (0.999)(0.02) + (0.01)(0.98) = 0.02978.
Test B: P(T+) = (0.99)(0.02) + (0.001)(0.98) = 0.02078.
(b) P(H+|T+) = P(H+,T +)
P(T+) =P(H+)P(T+|H+)
P(T+) .
Test A: P(H+|T+) = (0.02)(0.999)
(0.02978) = 0.67092.
Test B: P(H+|T+) = (0.02)(0.99)
(0.02078) = 0.952839.
(c) P(H+|T−) = P(H+)P(T−|H+)
P(T−)=P(H+)[1−P(T+|H+)]
[1−P(T+)] .
Test A: P(H+|T−) = (0.98)(1−0.999)
(1−0.2978) = 0.00101.
Test B: P(H+|T−) = (0.98)(1−0.99)
(1−0.2078) = 0.010008.
3. (a) Note that each row of the matrix (call it A) is just the conditional pmf of Yconditioned on
the value of X. Let ~
PX= [0.5,0.25,0.25] be the pmf vector for X. Then, ~
PY=~
PXA. More
specifically, P{Y =j}=P3
i=1 P{Y =j|X =i}P{X =i}
=
0.8×0.5+0.05 ×0.25 + 0.15 ×0.25 = 0.45 for j= 1
0.1×0.5+0.9×0.25 + 0.05 ×0.25 = 0.2875 for j= 2
0.1×0.5+0.05 ×0.25 + 0.8×0.25 = 0.2625 for j= 3
Sanity check: the three probabilities sum to 1.
(b) P{X =i|Y = 3}=P{Y = 3|X =i}P{X =i}
P{Y = 3}=
0.1×0.5
0.2625 =4
21 for i= 1,
0.05×0.25
0.2625 =1
21 for i= 2,
0.8×0.25
0.2625 =16
21 for i= 3.
This is the conditional pmf of Xgiven that Ywas observed to have value 3. Note that the sum
of the three probabilities is 1, as it should be for a valid pmf.
4. (a) P(F1) = P(F1|E1)P(E1) + P(F1|E2)P(E2) + P(F1|E3)P(E3) + P(F1|E4)P(E4)
= (0.9)(0.4) + (0)(0.3) + (0)(0.2) + (0)(0.1) = 0.36
(b) If the MS was found on the first page, it implies that the MS is located in cell 1 ⇒P(E1|F1) = 1.
(c) No matter which cell the MS is located in, the first time the MS is paged in the cell in which
it is located, it will be missed with probability 0.1. Given that the MS is missed in the first
round, it will be missed again in the second round with conditional probability 0.1. So P(G8) =
(0.1)(0.1) = 0.01.
(d) Now P(F2|G1) = P(F2G1
P(G1). By the law of total probability with the partition consisting of events
E1, E2, E3, E4,
P(G1) = P(G1|E1)P(E1) + P(G1|E2)P(E2) + P(G1|E3)P(E3) + P(G1|E4)P(E4)
= (0.1)(0.4) + (1)(0.3) + (1)(0.2) + (1)(0.1) = 0.64
and
P(F2G1) = P(F2G1|E1)P(E1) + P(F2G1|E2)P(E2) + P(F2G1|E3)P(E3)
+P(F2G1|E4)P(E4)
= (0)(0.4) + (0.9)(0.3) + (0)(0.2) + (0)(0.1) = 0.27
