University Solutions to Second Hour Exam ECE 313
of Illinois Spring 2002
1. Let A, B, and C respectively denote the events that the U-O, O-C, and U-C links are in working condition.
The capacities are marked on the Karnaugh map below (in which each cell has probability 1/8). It is easily
seen that Z takes on values 0, 56, 100, and 156 with probabilities 3/8, 1/8, 3/8, and 1/8 respectively, and
hence E[Z] = 0×3
8 + 56×1
8 + 100×3
8 + 156×1
8 = 0 + 56 + 300 + 156
8 = 512
8 = 64.
B
A
C
100 100 100
156
560 0 0
2
11/2
1
2. 1 = ∫
–∞
∞
fX(u)du = ∫
0
1
a + bu du = au + bu2/2
1
0= a + b/2.
Also, 2
3 = E[X] = ∫
–∞
∞
ufX(u)du = ∫
0
1
au + bu2 du = au2/2 + bu3/3
1
0= a/2 + b/3. Hence, a = 0, b = 2. The
pdf is thus 2u on [0, 1]. P{X < 1/2} = ∫
0
1/2
2u du = 1
4. This is even easier if you sketch the pdf as shown.
3.(a) The pdf of X has value 0.2 for –1 ≤ u ≤ 4, is as shown below in the left-hand figure below. We have that
E[Y] = E[|X–2|] = ∫
–1
2 0.2(2–u) du + ∫
2
4
0.2(u–2) du = 0.2(2u – u2/2)
2
–1 + 0.2(u2/2–2u)
4
2 = 1.3.
0.2
–1 4
v
–1
v
2
1
2
2+v2–v 4
2–v
(b) Y takes on values in the range [0, 3]. From the middle and right-hand figures above, we see that
for any v, 0 ≤ v ≤ 2, FY(v) = P{Y ≤ v} = P{2–v ≤ X ≤ 2+v} = 0.2(2+v–(2–v)) = 0.4v, while
for any v, 2 ≤ v ≤ 3, FY(v) = P{2–v ≤ Y < 4} = 0.2(4–(2–v)) = 0.2(2+v).
Hence, fY(v) =
0.4, 0 ≤ v ≤ 2,
0.2, 2 ≤ v ≤ 3
0, elsewhere which is easily verified to be a valid pdf.
4.(a),(b) As studied in class, X is an exponential random variable with parameter µ. Hence, P{X > τ} = exp(–µτ)
and fX(u) = µ•exp(–µu) for u > 0, and 0 otherwise.
(c) P{X > τ | A} is obviously 0 if τ ≥ 3, while for 0 ≤ τ < 3,
P{X > τ | A} = P{No arrivals in (0,τ] | two arrivals in (0,3]}
= P{No arrivals in (0,τ] ∩ two arrivals in (0,3]}/P{two arrivals in (0,3]}
= P{No arrivals in (0,τ] ∩ two arrivals in (τ,3]}/P{two arrivals in (0,3]}
= [exp(–µτ)•exp(–µ(3–τ))•(µ(3–τ))2/2!]/exp(–3µ)•(3µ)2/2! = (3–τ)2/9.
(d) fX|A(u|A) = – d
dτ P{X > τ | A} = (6 – 2τ)/9, 0 ≤ τ < 3, and 0 otherwise.
5. P{{|X – 4| > 3} = P{X > 7} + P{X < 1} = 1 – Φ( )
7–2
5 + Φ( )
1–2
5 = 1 – Φ(1) + Φ(–0.2)
= 1 – Φ(1) + 1 – Φ(0.2) = 2 – 0.8413 – 0.5793 = 0.5794.
P{X < 3 | X > 2} = P{2 < X < 3}/P{X > 2} = 2[ ]
Φ( )
3–2
5 – Φ( )
2–2
5 = 2(Φ(0.2) – 0.5)
= 1.1586 – 1 = 0.1586