University of Illinois Fall 2008
ECE 313: Solutions to Problem Set 3
1. (a) Γ(1) = Z∞
0
exp(−t)dt =−exp(−t)
∞
0
= 1.
(b) For α > 0, Γ(α+ 1) = Z∞
0
tαexp(−t)dt =tα[−exp(−t)]
∞
0−Z∞
0
αtα−1[−exp(−t)] dx
= 0 + αZ∞
0
tα−1exp(−t)dt =αΓ(α).
Hence Γ(n+ 1) = nΓ(n) = n(n−1)Γ(n−1) = ··· =n(n−1) ···2·1·Γ(1) = n!.
(c) Similarly, Γ(α+ 1) = αΓ(α) = α(α−1)Γ(α−1) = ··· =α(α−1)(α−2) ···Γ(α−bαc) where bαc
is the integer part of α, and 0 < α − bαc<1.
(d) Using the substitution x=√2t,dt/√t=√2dx and the suggested change to polar coordinates,
Γ1
2=Z∞
0
t−1/2exp(−t)dt =√2Z∞
0
exp(−x2/2) dx =√2Z∞
0
exp(−y2/2) dy
=2Z∞
0Z∞
0
exp(−[x2+y2]/2) dx dy1/2
= 2Z∞
r=0 Zπ/2
θ=0
exp(−r2/2) ·r dθ dr!1/2
=πZ∞
r=0
rexp(−r2/2)dr1/2
=√πvia the result of Problem 5(b) of Problem Set 1.
2. (a) m+n
kcommittees of size kcan be drawn from a set of mmen and nwomen. Of these,
m
i n
k−icommittees have imen and k−iwomen. Summing over all possible values of i, we
conclude that m+n
k=
k
X
i=0 m
i n
k−i.
(b) The coefficient of xkin the polynomial (1 + x)m+nis m+n
k. Now, if h(x) = f(x)g(x),
then the coefficient hkof h(x) is the discrete convolution of the coefficients of f(x) and g(x):
viz., hk=Pifigk−i. Applying this to f(x) = (1 + x)mand g(x) = (1 + x)n, we get that
m+n
k=
k
X
i=0 m
i n
k−ias before.
(c) With m=k=n,2n
n=
n
X
i=0 n
i n
n−i=
n
X
i=0 n
i2
since n
n−i=n
i.
3. Consider the experiment of picking a person at random from the 100,000 population of the town, and
let A,B, and Cdenote the events that the person selected reads Newspapers I, II, and III resp ectively.
We are given that P(A)=0.1, P (B)=0.3,P (C)=0.05; P(AB)=0.08, P (AC)=0.02, P (BC ) =
0.04; and P(AB C) = 0.01. Marking these probabilities on a Karnaugh map, we readily deduce
the probabilities shown in the figure on the next page. (Venn diagramists and those who believe in
brainpower alone unaided by any diagrams can work it out that P(AB) = P(AB C) + P(ABCc) can
be used to deduce that P(ABC c) = 0.07 etc.) From the diagram, it is easy to read off that
(a) P{read one newspaper only}=P(ABcCc) + P(AcB Cc) + P(AcBcC) = 0.2⇒20,000 people
read only one newspaper.
(b) P{read at least two newspapers}=P(AB ∪AC ∪BC) = 0.12 ⇒12,000 people read at least two
newspaper.
(c) P{read at least one of I and III and also read II}=P((A∪C)∩B) = 0.11 ⇒11,000 people read
at least one of I and III and also read II.
