ECE 534 Random Processes Instructor: R. Srikant
Probability Review Quiz
Each problem is worth 20 points. There are five problems.
1. The number of ways in which you can be dealt 5 cards is 52
5. There are 4 ways to choose a suit,
and having chosen a suit there are 13
5ways of choosing 5 cards. Therefore there are 4 13
5ways in
which you can be dealt 5 cards from the same suit. The required probability therefore is
413
5
52
5.
Another way to solve this problem is to note that the first card that is dealt can be any suit, the probability
that the second card is the same suit as the first card is 12/51,the probability that the third card is also
the same suit as the first card is 11/50,and so on. Thus, the required probability is
11 ∗10 ∗9∗8
51 ∗50 ∗49 ∗48 .
2 (a). The z−transform is given by
ΨX(z) =
∞
X
i=0
p(i)zi(1)
=
∞
X
i=0
λie−λ
i!zi(2)
=e−λ
∞
X
i=0
(λz)i
i!(3)
= exp(λ(z−1)).(4)
(b).
ΨX1+X2(z) = E(zX1+X2) = E(zX1)E(zX2) = ΨX1(z)ΨX2(z) = exp((λ1+λ2)(z−1)),
where the second equality above follows from the independence of X1and X2.Thus, X1+X2is also Poisson
with mean λ1+λ2.
3. As shown in Figure 1, Y=X2−X1above the line x1=x2, and Y=X1−X2below the line x1=x2.
The c.d.f. of Yby definition is FY(y) = Pr[Y≤y]. The area corresponding to {Y≤y}in the (X1, X2)
plane is the region between the lines x2=x1+yand x2=x1−ycontained in the square [0,1] ×[0,1]. This
is easily computed to be y(2 −y). So
FY(y) = y(2 −y).
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