UNIVERSITY OF ILLINOIS, URBANA-CHAMPAIGN
Department of Electrical and Computer Engineering
ECE 313: Probability with Engineering Applications-Fall 2003
Problem Set 2 Solution
Problems to be turned in:
1. (a) Expanding (1 + x)3and (1 + x)4:
(1 + x)3=(1+x)2(1 + x)
=(1+2x+x2)(1 + x)
=1+2x+x2+x+2x2+x3
=1+ 3
(3
1)=3
x+3
(3
2)=3
x2+x3.
(1 + x)4=1+3x+3x2+x3+x+3x2+3x3+x4
=1+ 4
(4
1)=4
x+6
(4
2)=6
x2+4
(4
3)=4
x3+x4.
(b) There are two possible answers:
(i) n
kis the coefficient of xkin the Taylor series expansion of (1 + x)n,and
(ii) n
kdenotes the number of subsets of size kfrom a set of nobjects.
Answer (ii) is the reason why n
kis called the binominal coefficient !
2. P(E⊕F)=P(E\F)+P(F\E)=P(E)+P(F)−2P(EF).
3. Since 1 ≥P(E∪F)=P(E)+P(F)−P(EF), P(EF )≥P(E)+P(F)−1.
4.Axioms1and2aresatisfied: (i)0≤f(E)≤1, and (ii) f(S) = 1. The only thing to check is
Axiom 3. Suppose Eiare disjoint sets,
f∞
i=1
Ei=
∞
i=1
f(Ei).
This is equivalent to saying that the number of times the outcome lies in ∞
i=1 Eiis equal to the
sum over iof the number of times the outcome lies in Ei.
5. (a) Since P(AB)+P(BC)+P(AC )−2P(ABC)=0.1+3P(BC)−0.1=P(AB∪BC ∪AC )=0.3,
P(BC)=0.1, and P(AC )=2P(BC)=0.2. The Venn diagram of the events A,B,andC
are shown below.
(b) From the Venn diagram, P(ACBCBC)=1−P(A∪B∪C)=1−0.45 = 0.55.
(c) P(ABCCC)=P(A)−P(AB ∪AC )=0.3−0.25 = 0.05. Similarly, P(ACBCC)=P(B)−
P(AB ∪BC)=0.05, and P(ACBCC)=P(C)−P(AC ∪BC)=0.05.
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