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Problem 1. Prove that for all n ∈ N, the following identity holds
Solution. By induction on n ≥ 1, with induction hypothesis
for all n ∈ N
Base case (n = 1):
Inductive step: Assume P(n), we need to show that P(n + 1) holds.
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as required.
We have shown that P(n) ⇒ P(n + 1). Thus, P(n) is true for all n ∈ N
Problem 2. Coin-Flip is a 2 player game. Each player wins with probability exactly 0.5. There
are no ties.
n people are playing a Coin-Flip tournament. Every person plays a Coin-Flip game
with every other person exactly once. Thus everybody plays n − 1 games. The outcomes of all
the games are mutually independent of one another.

We say that the tournament is a success if for every i ∈ {0, 1, . . . , n − 1}, there is exactly
one player, which we will refer to as pi, with exactly i wins.
(a) Prove that if the tournament is a success, then for any integers j, k with 0 ≤ k < j ≤ n
− 1, pj defeats pk.
(b) What is the probability that the tournament will be a success?

c) Show that your answer to part (b) is o(1). Solution. We have,
which for sufficiently large n, is clearly less than any positive constant, and thus is o(1).

Problem 3. A person is passing time by advancing a token on the set of natural numbers. In
the beginning, a token is placed on 0.
The person keeps playing moves forever. Each move proceeds as follows:
1. First the person tosses a fair coin (with heads/tails equally likely). 2
2. . Suppose the token is currently placed on n. If heads came up, then the person moves
the token to n + 3, otherwise he moves the token to n + 4.
For each n ∈ N, let En be the event ”There was a move on which the token landed on n”. Let pn
= Pr[En].
Find a recurrence relation for pn. You do not need to solve the recurrence, but you should
specify the boundary conditions that would be necessary to find a solution to the recurrence.

Problem 4. Exactly 1/5th of the people in a town have Beaver Fever©.
There are two tests for Beaver Fever, TEST1 and TEST2. When a person goes to a doctor
to test for Beaver Fever, with probability 2/3 the doctor conducts TEST1 on him and with
probability 1/3 the doctor conducts TEST2 on him.
When TEST1 is done on a person, the outcome is as follows:
• If the person has the disease, the result is positive with probability 3/4.
• If the person does not have the disease, the result is positive with probability 1/4.
When TEST2 is done on a person, the outcome is as follows:
• If the person has the disease, the result is positive with probability 1.
• If the person does not have the disease, the result is positive with probability 1/2.
A person is picked uniformly at random from the town and is sent to a doctor to test for Beaver
Fever. The result comes out positive. What is the probability that the person has the disease?

Solution. Let B be the event that the person has BLAH. Let T1 be the event that the person is
tested with test1. Let T2 be the event that the person is tested with test2. Let P be the event
that the test comes out positive.
A tree diagram is worked out below with the given information:

The probability that a person has BLAH, given that the test comes out positive is:
Problem 5. Two identical complete decks of cards, each with 52 cards, have been mixed
together. A hand of 5 cards is picked uniformly at random from amongst all subsets of exactly 5
cards.

(a) What is the probability that the hand has no identical cards (i.e., cards with the same suit and
value. For example, the hand �Q♥, 5♠, 6♠, 8♣, Q♥� has identical cards.)? We can calculate this
probability by computing
There are 104 cards. There are 5 cards in a hand. Order does not matter. The total number of
possible hands is:
There are 52 possible card faces, and we can choose 5 of them if no identical cards are allowed.
Additionally, each card can be from either deck 1 or deck 2. Therefore the number of hands
with no identical cards, chosen from 2 decks is:
Therefore the probability of drawing a hand with no identical cards is:

(b) What is the probability that the hand has exactly one pair of identical cards? This can be
solved by a similar approach. A hand of this type can be distinguished
by the face (suit and value) of the repeated card, and by the faces of the 3 non-repeated
cards. There are 52 possible values for the face of the repeated card.
Problem 6. [28 points] Scores for a final exam are given by picking an integer uniformly at
random from the set {50, 51, . . . , 97, 98}. The scores of all 128 students in the class are
assigned in this manner. For parts (a), (b), (c) and (d) you may NOT assume that these scores
are assigned independently. For parts (e), (f), (g) and (h) you MAY assume that these scores
are assigned independently.

(a) For i ∈ {1, . . . , 128}, what is E[Si] ?
(b) Show that E[S] = 74. Make no independence assumptions.
(c) Prove that
Make no independence assumptions.
(d) Improve your previous bound by using the fact that the minimum possible score is 50.
Prove that
Make no independence assumptions.
(e) For the remaining problems, assume that all the scores are assigned mutually
independently. Use Problem 1 of this final to find V ar[Si].

(f) What is V ar[S]?
(g) What is the standard deviation of S?
(h) Prove, using the Chebyshev Inequality, that
Solution.
(a) We simply take the average of the numbers from 50 to 98. Thus, E[Si] = 74.
(b) By linearity of expectation,
(c) By Markov’s inequality,

(d) We define a random variable T = S − 50, and thus E[T] = E[S] − 50 = 24. Now we just
apply Markov’s inequality:
(e) We define Ti = Si − 50.
(g) The standard deviation of S is simply the square root of the variance of S:

(h) Using Chebyshev’s inequality,
Problem 7. 1000 files F1, F2, . . . , F1000 have just reached a disk manager for writing onto
disk. Each file’s size is between 0MB and 1MB. The sum of all files’ sizes is 400MB.
The disk manager has 4 disks under its control. For each file Fi, the disk manager chooses
a disk uniformly at random from amongst the 4 disks, and Fi is written to that disk. The
choices of disk for the different files are mutually independent.
(a) What is the expected number of files that will be written to the first disk?
We can use indicator variables. For each file, Pi = 1 if Fi is written to the first disk. The
chance of an individual file being written to the first disk is 1/4. By linearity of
expectation, the expected number of files written to the first disk is the sum of the expected
values of Pi’s. The expected value of each indicator variable is 1/4, and
so the expected number of files to be written to the first disk

(b) What is the expected number of bytes written on the first disk?
We can say that each file Fi has bit size Si. Each file has a 1/4 chance of being written do the
first disk. Therefore, by linearity of expectation, the expected number of bytes written to the
first disk is the sum of the expected number of bytes per file written to the first disk, which
is:
(c) Find the best upper bound you can on the probability that 200MB or more are written on
the first disk?
For this we can use the first Chernoff bound, which is:
The Chernoff bound only works if X is the sum of random variables that each take on a value
between 0 and 1. The file size of each file in the first disk is between 0 and 1Mb . So we can
define X to be the total number of bytes in disk 1. The expected value of X is 100, so we take c
to be 2. We get:

(d) Find the best upper bound you can on the probability that there is some disk with 200MB or
more written on it?
For this we can use the Union Bound along with our result from above. The probability of
this event happening in one or more disks is upper bounded by the sum of the probabilities of
the event happening in each disk. This gives us an upper bound of

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