matt grime said:
To make this rgorous you wuld need a uniform probability measure on the natrual numbers. none such exists.
however, the next best thing is this: suppose that you pick at random with uniform prob from the set {1,..,N} what are the chances.
Notice the redundancy of your requirements: if a number is not divisible by 3 it is certainly not divisible by 6.
roughly 1/2 are divisible by 2, 1/5 by 5 and how many by both? now think about those that are not divisible be 3 and 4.
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So, I am going to show as I answered to this question.
Consider the events and \Omega = {1,2,3, ..., 120}, natural numbers:
A = {integers which are divisible by 3}, with P(A) = 1/3;
B = {integers which are divisible by 4}, with P(B) = 1/4;
C = {integers which are divisible by 6}, with P(C) = 1/6;
D = {integers which are divisible by 2}, with P(D) = 1/2;
E = {integers which are divisible by 5}, with P(E) = 1/5.
I need to find the probability of the set: A^cB^cC^c(D \cup E).
Let A^cB^cC^c(D \cup E) = (A^cB^cC^cD) \cup (A^cB^cC^cE).
and
(A^cB^cC^cD) \cap (A^cB^cC^cE) = (A^cB^cC^cDE) = DE - ABCDE.
then
P(A^cB^cC^cDE) = P(DE) - P(ABCDE) = 1/10 - 1/60 = 1/12.
P(DE) = 1/10;
P(ABCDE) = 1/60.
and
P(A^cB^cC^cD) = P(D) - P(ABCD) = 1/2 - 1/12 = 5/12.
P(A^cB^cC^cE) = P(E) - P(ABCE) = 1/5 - 1/60 = 11/60.
=>
(A^cB^cC^cD) \cup (A^cB^cC^cE) = P(A^cB^cC^cD) + P(A^cB^cC^cE) - P(A^cB^cC^cDE) = 5/12 + 11/60 - 1/12 = 31/60
However, the right answer is 14/60. I committed some error? Where I can have wrong