ECE 534: Random Pro cesses Spring 2010
Solutions to Problem Set 1
1.1 Simple events
(a) Ω = {0,1}8, or Ω = {x1x2x3x4x5x6x7x8:xi∈ {0,1}for each i}. It is natural to let F
be the set of all subsets of Ω. Finally, let P(A) = |A|
256 , where |A|denotes the cardinality
of a set |A|.
(b) E1={01010101,10101010}and P[E1] = 2
256 =1
128 .
E2={00110011,01100110,11001100,10011001}and P[E2]=4/256 = 1/64.
E3={x∈Ω : x1+· · · +x8= 4}and P[E2] = 8
4/256 = 70/256 = 35/128.
E4={11111111,11111110,11111101,10111111,01111111,00111111,01111110,11111100}
and P[E4] = 8/256 = 1/32.
(c) E1⊂E3, so P[E1|E3] = |E1|/|E3|= 2/70 = 1/35.
E2⊂E3, so P[E2|E3] = |E2|/|E3|= 4/70 = 2/35.
1.5 Conditional probability of failed device given failed attempts
(a) P[first attempt fails]= 0.2 + (0.8)(0.1) = 0.28.
(b) P[server is working |first attempt fails] =
P[server working, first attempt fails]/P[first attempt fails] = (0.8)(0.1)/0.28 ≈0.286.
(c) P[second attempt fails |first attempt fails] =P[first two attempts fail]/P[first attempt
fails] = [0.2 + (0.8)(0.1)2]/0.28 ≈0.743.
(d) P[server is working |first and second attempts fail ] =P[server is working and first two
attempts fail]/P[first two attempts fail] = (0.8)(0.1)2/[0.2 + (0.8)(0.1)2]≈0.0385.
1.11 A CDF of mixed type
(a) FX(0.8) = 0.5.
(b) There is a half unit of probability mass at zero and a density of value 0.5 between 1
and 2. Thus, E[X]=0×0.5 + R2
1x(0.5)dx = 3/4 and,
(c) E[X2]=02×0.5 + R2
1x2(0.5)dx = 7/6.So Var(X)=7/6−(3/4)2= 29/48.
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