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Problem 1. Give an inductive proof that the Fibonacci numbers Fn and
Fn+1
are relatively prime for all n ≥ 0. The Fibonacci numbers are deﬁned as
follows:
F0 = 0 F1 =
1
Fn = Fn−1
+ Fn−2
(for n ≥ 2)
Solution. We use induction on n. Let P (n) be the proposition that Fn and
Fn+1 are relatively prime.
Base case: P (0) is true because F0 = 0 and F1 = 1 are relatively prime.
Inductive step: Assume that P (n) is true where n ≥ 0; that is, Fn and Fn+1
are relatively prime. We must show that Fn+1 and Fn+2 are relatively
prime as well. If Fn+1 and Fn+2 had a common divisor d > 1, then d
would also divide the linear combination Fn+2 − Fn+1 = Fn,
contradicting the assumption that Fn and Fn+1 are relatively prime. So
Fn+1 and Fn+2 are relatively prime.
The theorem follows by induction.
Problem 2. The double of a graph G consists of two copies of G with
edges joining corresponding vertices. For example, a graph appears
below on the left and its double appears on the right.
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Some edges in the graph on the right are dashed to clarify its
structure.
(a) Draw the double of the graph shown below.
Solution.
(b) Suppose that G1 is a bipartite graph, G2 is the double of G1, G3 is the
double of
G2, and so forth. Use induction on n to prove that Gn is bipartite for all n ≥ 1.
Solution. We use induction. Let P (n) be the proposition that Gn is
bipartite.
Base case: P (1) is true because G1 is bipartite by assumption.
Inductive step: For n ≥ 1, we assume P (n) in order to prove P (n + 1).
The graph Gn+1 consists of two subgraphs isomorphic to Gn with edges
joining corresponding vertices. Remove these extra edges. By the
assumption P (n), we can color each vertex of one subgraph black or
white so that adjacent vertices get different colors.

If we color the corresponding vertices in the other subgraph oppositely,
then adjacent vertices get different colors within that subgraph as well.
And now if we add back the extra edges, each of these joins a white
vertex and a black vertex. Therefore, Gn+1 is bipartite.
The theorem follows from the principle of induction.
Problem 3. Finalphobia is a rare disease in which the victim
has the delusion that he or she is being subjected to an intense
mathematical examination.
• A person selected uniformly at random has finalphobia
with probability 1/100.
• A person with finalphobia has shaky hands with
probability 9/10.
• A person without finalphobia has shaky hands with
probability 1/20.
What is the probablility that a person selected uniformly at
random has finalphobia, given that he or she has shaky hands?
Solution. Let F be the event that the randomly-selected
person has finalphobia, and let S be the event that he or she has
shaky hands. A tree diagram is worked out below:

n
o
ye
s
99/10
0
yes
1/10
0
has
shaky
hands?
yes
9/1
0
1/1
0
no
1/20
19/2
0
no
has
finalphobia?
F S
X X 9/100
0
1801/20
00
99/200
0
1/100
0
Probabili
ty
X
X
The probability that a person has ﬁnalphobia, given that he or she
has shaky hands is:
Pr (F
∩S)
Pr
(S)
9/10
00
Pr (F |
S)
=
=
9/1000 + 99/2000
18
=
18 + 99
18
117
=

Problem 4. Suppose that you roll five 6-sided dice that are fair and
mutually independent. For the problems below, answers alone are
sufficient, but we can award par tial credit only if you show your work.
Also, you do not need to simplify your answers; you may leave
factorials, binomial coefficients, and arithmetic expressions unevaluated.
(a) What is the probability that all five dice show
different values? Example: (1, 2, 3, 4, 5) is a roll of this
type, but (1, 1, 2, 3, 4) is not.
Solution. The probability space is the uniform distribution on the 65
possible num bers rolled on the five (distinguishable) dice. So the
probability that all dice are different is the number of outcomes in
which the dice have distinct values divided by 65. There are (6)5 =
6 ·5 ·4 ·3 ·2 such outcomes so
Pr (all rolls distinct) = 6 ·5 ·4 ·3 ·2
= 5!
An alternative approach uses the observation that the conditional
probability that the i + 1st die value differs from the preceding
rolls, given that the first i values differ, is (6 − i)/6 for 1 ≤ i ≤ 4, and
the probability that all five values are different is the product of
these conditional probabilities, namely
65 64
Pr (all rolls distinct) = - - - -
-
5 4 3 2 5!
6· 6 · 6 · 6 64
(b) What is the probability that two dice show the same value and
the remaining three dice all show different values?
Example: (6, 1, 6, 2, 3) is a roll of this type, but (1, 1, 2, 2, 3) and (4, 4,
4, 5, 6) are not.

Solution. There are 5 2 possible pairs of rolls that might have the same
value and 6 possibilities for what this value is. There 5 · 4 · 3 possible
distinct values for the remaining three rolls. So
Pr(exactly two values the same) =
An alternative way to count is: there are 4 sets of four values among the
five dice, 4 choices for which of these values is repeated, and by the
Bookkeeper rule, 5 = 5!/2 permutations of a sequence consisting of five
values, one of which 2,1,1,1 appears twice. So,
Pr(exactly two values the same) =
(c) What is the probability that two dice show one value, two different
dice show a second value, and the remaining die shows a third value?
Example: (6, 1, 2, 1, 2) is a roll of this type, but (4, 4, 4, 4, 5) and (5, 5,
5, 6, 6) are not.
2
Solution. There are sets of two values that might be duplicated. There are
rolls where larger duplicated value may come up and (3/2) remaining rolls
where the
smaller duplicated value may come up. There is only 1 remaining roll
where the nonduplicated value may then come up, and 4 remaining
values it could take. So,

Pr(exactly two pairs of same values) =
Alternatively, there are sets of three values among the five dice, 3 choices
for 3 which of these values is not repeated, and by the Bookkeeper rule,
2,2 5 ,1 permutations of a sequence consisting of five values, two of
which appear twice. So,
Pr(exactly two pairs of same values) =
Problem 5. An electronic toy displays a 4 × 4 grid of colored squares.
At all times, four are red, four are green, four are blue, and four are
yellow. For example, here is one possible configuration:
For parts (a) and (b) below, you need not simplify your answers.
(a) How many such configurations are possible?
Solution. This is equal to the number of sequences containing 4 R’s, 4 G’s,
4 B’s, and 4 Y’s, which is:

For parts (a) and (b) below, you need not simplify your answers.
(a)How many such configurations are possible?
Solution. This is equal to the number of sequences containing 4 R’s, 4
G’s, 4 B’s, and 4 Y’s, which is:
16!
(4!)4
(b)Below the display, there are five buttons numbered 1, 2, 3, 4, and 5.
The player may press a sequence of buttons; however, the same button
can not be pressed twice in a row. How many different sequences of n
button-presses are possible?
Solution. There are 5 choices for the first button press and 4 for each
subsequence press. Therefore, the number of different sequences of n
button presses is 5 ·4n−1.
(c)Each button press scrambles the colored squares in a complicated,
but nonran dom way. Prove that there exist two different sequences of
32 button presses that both produce the same configuration, if the
puzzle is initially in the state shown above. (Hint: 432 = 1616 > 16!)
Solution. We use the Pigeonhole Principle. Let A be the set of all
sequences of 32 button presses,

let B be the set of all configurations, and let f : A → B map each
sequence of button presses to the configuration that results. Now:
32
|A| > 4 > 16! > |B|
Thus, by the Pigeonhole Principle, f is not injective; that is, there
exist distinct el ements a1, a2 ∈ A such taht f (a1) = f (a2). In
other words, there are two different sequences of button presses
that produce the same configuration.
Problem 6. MIT students sometimes delay laundry for a few days.
Assume all random values described below are mutually
independent.
(a) A busy student must complete 3 problem sets before doing
laundry. Each prob lem set requires 1 day with probability 2/3
and 2 days with probability 1/3. Let B be the number of days a
busy student delays laundry. What is Ex (B)?
Example: If the first problem set requires 1 day and the second
and third problem sets each require 2 days, then the student
delays for B = 5 days.
Solution. The expected time to complete a problem set is:
Therefore, the expected time to complete all three problem sets is:

(b) A relaxed student rolls a fair, 6-sided die in the morning. If he rolls a
1, then he does his laundry immediately (with zero days of delay).
Otherwise, he delays for one day and repeats the experiment the
following morning. Let R be the number of days a relaxed student
delays laundry. What is Ex (R)?
Example: If the student rolls a 2 the ﬁrst morning, a 5 the second
morning, and a 1 the third morning, then he delays for R = 2 days.
Solution. If we regard doing laundry as a failure, then the mean time to
failure is 1/(1/6) = 6. However, this counts the day laundry is done, so
the number of days delay is 6 − 1 = 5. Alternatively, we could derive
the answer as follows:
Ex (B) = Ex (pset1) + Ex (pset2) + Ex (pset3)

(c) Before doing laundry, an unlucky student must recover from illness
for a number of days equal to the product of the numbers rolled on
two fair, 6-sided dice. Let U be the expected number of days an
unlucky student delays laundry. What is Ex (U )?
Example: If the rolls are 5 and 3, then the student delays for U = 15
days.
Solution. Let D1 and D2 be the two die rolls. Recall that a die roll
has expectation
7/2. Thus:
(d) A student is busy with probability 1/2, relaxed with probability 1/3,
and unlucky with probability 1/6. Let D be the number of days the
student delays laundry. What is Ex (D)? Leave your answer in terms
of Ex (B), Ex (R), and Ex (U ).
Solution.
Final Exam
Problem 7. I have twelve
cards:

I shufﬂe them and deal them in a row. For example, I might get:
What is the expected number of adjacent pairs with the same value? In the
example, there are two adjacent pairs with the same value, the 3’s and the
5’s.
We can award partial credit only if you show your work.
Solution. Consider an adjacent pair. The left card matches only one of
the other 11 cards, which is equally likely to be in any of the 11 other
positions. Therefore, the proba bility that an adjacent pair matches is 1/11.
Since there are 11 adjacent pairs, the expected number of matches is 11 ·
1/11 = 1 by linearity of expectation.
Problem 8.Each time a baseball player bats, he hits the ball with some
proba bility. The table below gives the hit probability and number of
chances to bat next season
for ﬁve players.
player prob. of
hit
# chances
to bat
Player A 1/3 300
Player B 1/4 200
Player C 1/4 400
Player D 1/5 250
Player E 2/5 500

(a) Let X be the total number times these ﬁve players hit the ball next
season. What is Ex (X)?
Solution. The expected number of events that happen is equal to the
sum of the event probabilities. So we have:
(b) Give a nontrivial upper bound on Pr (X ≥ 1500) and justify your
answer. Do not
assume that hits happen mutually independently.
Solution. We use the Markov Inequality:
(c) Using a Chernoff inequality, give a nontrivial upper bound on Pr (X ≤
400). For
this part, you may assume that all hits happen mutually independently.
Solution. We use the Chernoff Inequality:
This gives:

This gives:

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