Probability of Misprints in Documents and Chernoff Bounds - Prof. Samir Khuller, Assignments of Computer Science

Download Probability of Misprints in Documents and Chernoff Bounds - Prof. Samir Khuller and more Assignments Computer Science in PDF only on Docsity! CMSC 498K : Homework 2 Noah Luck Easterly March 6, 2008 Problem 1. Suppose that a document contains 300 pages and contains 20 misprints. What is the probability that there is more than one misprint on a particular page? Assuming that all the misprints are independently randomly distributed, we have Prob( misprint i is on page j ) = 1 300 Prob( page j has exactly k misprints ) = ( 20 k ) ( 1 300 )k (299 300 )20−k Prob( page j has m or more misprints ) = 20∑ k=m Prob( page j has exactly k misprints ) = 1− m−1∑ k=1 Prob( page j has exactly k misprints ) ⇒ Prob( page j has more than one misprint ) = 1− [Prob( page j has exactly 0 misprints ) + Prob( page j has exactly 1 misprint )] = 1− [( 20 0 ) ( 1 300 )0 (299 300 )20 + ( 20 1 ) ( 1 300 )1 (299 300 )19] = 1− [( 299 300 )20 + 20 ∗ 299 19 30020 ] = 1− 299 19 30020 [299 + 20] = 1− 299 19 30020 ∗ 319 ≈ 2.03 ∗ 10−3 Problem 2. Suppose 4% of the parts made in a factory are defective. Suppose we ship out a batch of 25 parts. What is the probability that there are no defective parts in this batch? Assuming that all occurrences of parts are independent, we have Prob( any particular part is defective ) = 1 25 1 ⇒ Prob( in a batch of n parts, exactly k are defective ) = ( n k ) ( 1 25 )k (24 25 )n−k Prob( in a batch of 25 parts, exactly 0 are defective ) = ( 25 0 ) ( 1 25 )0 (24 25 )25 = ( 24 25 )25 ≈ 0.360 Problem 3. Suppose we toss 1000 fair coins. Use Chernoff bounds to derive an upper bound on the proba- bility that we either get less than 400 heads or more than 600 heads is small. A fair coin means that Prob(head) = Prob(tail) = 1 2 Each fair coin is independent. Define 1000 random variables Xi = { 1, if the ith coin is a head; 0, otherwise. Then Prob(Xi = 1) = 1 2 Let X = 1000∑ i=1 Xi Then µ = E[X] = E[ 1000∑ i=1 Xi] = 1000∑ i=1 E[Xi] = 1000∑ i=1 1 2 = 500 So Prob(less than 400 heads or more than 600 heads) = Prob(|X − µ| ≥ 100) The corollary to Chernoff’s bounds states: Prob(|X − µ| ≥ δµ) ≤ 2e(−µδ2)/2 So, with δ = 15 , we have, Prob(less than 400 heads or more than 600 heads) ≤ 2e(−500∗ 1 5 2 )/2 = 2e−10 ≈ 4.54 ∗ 10−5 2