Jul 24, 2020

The Binomial Distribution p(x = 3) = 4!3!1! × (1/6) 3 (5/6) 1 = 4 × (1/6) 3 × (5/6) = 0.0154 P(X = 4) = 4! 4!0! × (1/6) 4 (5/6) 0 = 1 × (1/6) 4 × 1 = 0.0008 Summary: "for the 4 throws, there is a 48% chance of no twos, 39% chance of 1 two, 12% chance of 2 twos, 1.5% chance of 3 twos, and a tiny 0.08% … Expected Value The expected value of a random variable XY(x,y), then =!! xy E(g(X,Y))g(x,y)p XY (x,y) If X and Y have a joint probability density function f XY(x,y), then !! " #" " #" E(g(X,Y))=g(x,y)f XY (x,y) It is important to note that if the function g(x,y) is only dependent on either x or y the formula above reverts to the 1-dimensional case. Ex. Suppose X and Y have a joint pdf f XY(x,y Shortlisted Problems with Solutions If x,y ∈ A (possibly x = y) then x2 +kxy +y2 ∈ A for every integer k. Determine all pairs m,n of nonzero integers such that the only admissible set containing both m and n is the set of all integers. N2. Find all triples (x,y,z) of positive integers such that x ≤ y ≤ z and x3(y3 +z3) = 2012(xyz +2). N3.

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