Math 213 Exam 2 Solutions: Problems and Answers, Exams of Discrete Mathematics

The solutions to exam 2 of math 213, including problems on probability, recurrence relations, and combinatorics. Students can use this document as a study resource to review concepts covered in the exam.

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2010/2011

Uploaded on 06/27/2011

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Math 213 Exam 2 9Solutions)
Prof. I.Kapovich April 4, 2008
Problem 1[20 points]
For each of the following statements indicate whether it is true or false.
You DO NOT have to explain your answers.
(1) For any random variables Xand Ywe have V(X+Y) = V X +V Y .
(2) For S={−5,3,1,5}, ordered as listed, the permutation 3,1,5,5
is a derangement.
(3) Let Abe the set of all bit-strings of length 10. Let Rbe a relation
on Aconsisting of all pairs (x, y)A×Asuch that the strings x
and yagree in their first two bits.
Then Ris an equivalence relation with exactly two equivalence
classes.
(4) Consider a recurrence relation an= 3an1+n.
Then for every c, α, β R,an=c3n+α+βn is a solution of the
above recurrence relation.
Answers:
(1) False. (The statement would be true if we knew that Xand Yare
independent random variables).
(2) True.
(3) False. (Although Ris an equivalence relation, it has four distinct
equivalence classes, corresponding to 4 distinct binary strings of length two
that can occur as initial segments of length two of strings from A).
(4) False. (One can find a particular solution of this recurrence relation
of the form an=α+βn, but the values of αand βthat work are not
arbitrary. In this case these specific values are α=3
4,β=1
2and the
general solution of the original recurrence relation is an=c3n3
41
2, where
cRis arbitrary.).
Problem 2[20 points]
Let n1 be an integer. A fair die is rolled ntimes, independently. Let
Z= 3X+ 2Ywhere Xis the number of tosses that came 1 up and where
Yis the number of tosses that came 6 up.
Find EZ and V Z .
Give all the details of your work.
Solution.
For i= 1,. . . , n let
Zi=(3,if the i-th toss is 1,
2,if the i-th toss is 6,0,otherwise.
1
pf3

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Math 213 Exam 2 9Solutions) Prof. I.Kapovich April 4, 2008 Problem 1[20 points] For each of the following statements indicate whether it is true or false. You DO NOT have to explain your answers.

(1) For any random variables X and Y we have V (X + Y ) = V X + V Y. (2) For S = {− 5 , 3 , 1 , 5 }, ordered as listed, the permutation 3, 1 , 5 , − 5 is a derangement. (3) Let A be the set of all bit-strings of length 10. Let R be a relation on A consisting of all pairs (x, y) ∈ A × A such that the strings x and y agree in their first two bits. Then R is an equivalence relation with exactly two equivalence classes. (4) Consider a recurrence relation an = 3an− 1 + n. Then for every c, α, β ∈ R, an = c 3 n^ + α + βn is a solution of the above recurrence relation.

Answers: (1) False. (The statement would be true if we knew that X and Y are independent random variables).

(2) True. (3) False. (Although R is an equivalence relation, it has four distinct equivalence classes, corresponding to 4 distinct binary strings of length two that can occur as initial segments of length two of strings from A).

(4) False. (One can find a particular solution of this recurrence relation of the form an = α + βn, but the values of α and β that work are not arbitrary. In this case these specific values are α = − 34 , β = − 12 and the

general solution of the original recurrence relation is an = c 3 n^ − 34 − 12 , where c ∈ R is arbitrary.).

Problem 2[20 points] Let n ≥ 1 be an integer. A fair die is rolled n times, independently. Let Z = 3X + 2Y where X is the number of tosses that came 1 up and where Y is the number of tosses that came 6 up. Find EZ and V Z. Give all the details of your work.

Solution. For i = 1,... , n let

Zi =

3 , if the i-th toss is 1, 2 , if the i-th toss is 6, 0 , otherwise. 1

2

Then Z = Z 1 +· · ·+Zn and the random variables Z 1 ,... , Zn are independent. For each i = 1,... , n we have

EZi = 3

and

E(Z i^2 ) = 3^2

+ 2^2

+ 0^2

Hence V Zi = E(Z^2 i ) − (EZi)^2 = 136 − 2536 = 5336. Since Z = Z 1 + · · · + Zn, it follows that

EZ = EZ 1 + · · · + EZn =

5 n 6 and, since Z 1 ,... , Zn are independent,

V Z = V Z 1 + · · · + V Zn =

53 n 36

Problem 3[20 points] Find the general solution of the recurrence relation

(†) an = 2an− 1 − an− 2 + 4.

Give all the details of your work.

Solution. The associated homogeneous relation is an = 2an− 1 − an− 2. Its charac- teristic equation is r^2 = 2r − 1 which factors as (r − 1)^2 = 0. Hence the homogeneous problem an = 2an− 1 − an− 2 has the general solution

a( nh )= α 1 n^ + βn 1 n^ = α + βn

where α, β ∈ R are arbitrary constants. Since 4 = 4 · 1 n, we should look for a particular solution of (†) of the form

a( np )= cn^2 ,

where c ∈ R. To find the specific value of c, we substitute a( np )= cn^2 into (†):

cn^2 = 2c(n − 1)^2 − c(n − 2)^2 + 4 cn^2 = 2cn^2 − 4 cn + 2c − cn^2 + 4cn − 4 c + 4 2 c = 4, c = 2.

Thus a( np )= 2n^2 is a particular solution of (†). Therefore the general solution of (†) is

an = a( nh )+ a( np )= α + βn + 2n^2 ,

where α, β ∈ R are arbitrary constants.

Problem 4[20 points]