Problems in Probability

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Fall 2011 Strategic Practice 3: Section 4 (Bernoulli and Binomial) - Question 3
Let X ~ Bin(n, p) and Y ~ Bin(m, p), independent of X.
(a) Show that X + Y ~ Bin(n + m, p), using a story proof.
(b) Show that X - Y is not Binomial.
(c) Find P(X = k|X + Y = j). How does this relate to the elk problem from HW 1?
Solution: (a) Interpret X as the number of successes in n independent Bernoulli trials and Y as the number of successes in m more independent Bernoulli trials, where each trial has probability p of success. Then X + Y is the number of successes in the n + m trials, so X + Y ~ Bin(n + m, p). (b) A Binomial can't be negative, but X - Y is negative with positive probability. (c) Using the definition of conditional probability to find P(X = k|X + Y = j) we get (nCk)(nC[j-k])/([n+m]Cj). This is exactly the same distribution as in the elk problem (it is called the Hypergeometric distribution). To see why, imagine that there are n male elk and m female elk, each of which is tagged with the word "success" with probability p (independently). Suppose we then want to know how many of the male elk are tagged, given that a total of j elk have been tagged. For this, p is no longer relevant, and we can "capture" the male elk and count how many are tagged, analogously to the original elk problem.
"Mathematics is the logic of certainty, but statistics is the logic of uncertainty."
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