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Suppose 120 vampires are at an undead convention. They have the following

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(a) How many vampires have none of the three restrictions? 

(b) If we select a vampire at random from among those who can't abide garlic, what is the probability that

they suff er from another restriction as well? 

71 shy away from holy symbols; 64 can't abide garlic;51 must sleep in their own coffins; 38 shy away from h.s. and can't abide garlic;36 shy away from h.s. and must sleep in their own coffs; 29 can't abide garlic and must sleep in their own coffs;17 suff er from all three restrictions.

Apr 11th, 2015

Let H be the set of all vampires that shy away from holy symbols, G be the set of all vampires that can't abide garlic, and C be the set of all vampires that must sleep in their own coffins. Then by H∩G we denote the set of all vampires that shy away from holy symbols and can't abide garlic, by G∩C the set of all vampires that can't abide garlic and must sleep in their own coffins, by H∩C the set of all vampires that shy away from holy symbols and must sleep in their own coffins, and by H∩G∩C the set of all vampires that suffer from all three restrictions. Denote by N(A) the number of elements in the set A. Let U be the universal set of all vampires at the convention.

Given N(U) = 120, N(H) = 71, N(G) = 64, N(C) = 51, N(H∩G) = 38, N(H∩C) = 36, N(G∩C) = 29, and N(H∩G∩C) = 17.

a) Denote the set of all vampires that suffer at least from one of the restrictions by HᴗGᴗC (the union of the three sets). Then N(HᴗGᴗC) = N(H) + N(G) + N(C) – N(H∩G) – N(H∩C) – N(G∩C) + N(H∩G∩C) = 71 + 64 + 51 – 38 – 36 – 29 + 17 = 100. Then the number of all vampires present at the convention and suffering from none of the three restrictions is 120 – 100 = 20.

b) In the set G we have the subsets H∩G and G∩C that consist of vampires suffering from two conditions. However, if we take N( H∩G) + N(G∩C) = 38 + 29 = 67, then the vampires that suffer from all three conditions will be counted twice. So, among all vampires that suffer from the condition G we have 38 + 29 – 17 = 50 vampires that suffer f rom another restriction as well. Finally, the probability of choosing a vampire from the set G that suffers from another restriction equals

50/64 = 25/32 = 0.78125.


Apr 11th, 2015

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