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What is the probability that Harry will go back home?

INSTRUCTIONS TO CANDIDATES
ANSWER ALL QUESTIONS

Problem 1. Harry is not sure if he wants to stay in his boarding school or go back home. To help him make up is mind, he wants to use his magic coin. When flipped, the magic coin comes up “heads” with probability 0.05 and “tails” with probability 0.95.

Harry decided to do the following; flip the magic coin 100 times and count the number of times it comes up “heads”. If this number is even, he will stay at the boarding school, and if this number is odd he will quit school and go back home.

What is the probability that Harry will go back home?

Clarification: Needless to say that your answer should be a “closed form” expression, that is, without any long summations/multiplications.

Solution:

Problem 2. Suppose n is odd. Let P([n]) denote the power set of [n], that is, the 2n subsets of [n]. We say that a family of sets F ⊆ P([n]) is nice if F does not contain 5 sets A, B, C, D, E satisfying A ⊆ B ⊆ C ⊆ D ⊆ E. Show that if F ⊆ P([n]) is nice then

|F| ≤ 2 n+ n.

(n − 1)/2 (n − 3)/2

In other words, a nice contains at most 2n(n−1)/2+ n(n−3)/2of the subsets of [n]. Also, find a nice ([n]) containing 2n(n−1)/2+ n(n−3)/2sets.

Solution:

Problem 3. Prove that if P is a set of 2018 distinct points in R2 then P determines at least 3 distinct (non-zero) distances. That is, there must be 3 pairs of points (a1, b1), (a2, b2), (a3, b3) (with a1 /= b1, a2 /= b2, a3 /= b3) so that d(a1, b1), d(a2, b2), d(a3, b3) are three distinct real numbers (where d(x, y) is the usual Euclidian distance in R2).

Solution:

Problem 4. Let p1(n) denote the number of ways we can express n as the sum of positive integers that are not divisible by 2018. Let p2(n) denote the number of ways we can express n as the sum of positive integers so that each integer appears at most 2017 times. Show that for every n we have p1(n) = p2(n).

Clarification: Needless to say that (as usual) order does not matter here. For example, p1(3) = 2, since {1, 2} is counted only once (rather than twice as (1, 2) (2, 1)).

Solution:

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