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Divisible of natural numbers

  • Thread starter Alexsandro
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Could Someone help with this question ?

What is the probability that a number of the set [itex]\Omega[/itex], first 120 natural numbers {1,2,3, ... , 120}, picked at random is not divisible by any of the number 3, 4, 6 but is divisible by 2 or 5 ?

Thanks
 
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Answers and Replies

matt grime
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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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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.
___________
So, I am going to show as I answered to this question.

Consider the events and [itex]\Omega[/itex] = {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: [itex]A^cB^cC^c(D \cup E)[/itex].
Let [itex]A^cB^cC^c(D \cup E)[/itex] = [itex](A^cB^cC^cD) \cup (A^cB^cC^cE)[/itex].

and

[itex](A^cB^cC^cD) \cap (A^cB^cC^cE) = (A^cB^cC^cDE) = DE - ABCDE.[/itex]

then

[itex]P(A^cB^cC^cDE)[/itex] = P(DE) - P(ABCDE) = 1/10 - 1/60 = 1/12.
P(DE) = 1/10;
P(ABCDE) = 1/60.

and

[itex]P(A^cB^cC^cD)[/itex] = P(D) - P(ABCD) = 1/2 - 1/12 = 5/12.
[itex]P(A^cB^cC^cE)[/itex] = P(E) - P(ABCE) = 1/5 - 1/60 = 11/60.

=>

[itex](A^cB^cC^cD) \cup (A^cB^cC^cE)[/itex] = [itex]P(A^cB^cC^cD)[/itex] + [itex]P(A^cB^cC^cE)[/itex] - [itex]P(A^cB^cC^cDE)[/itex] = 5/12 + 11/60 - 1/12 = 31/60

However, the right answer is 14/60. I committed some error? Where I can have wrong
 
Gokul43201
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Haven't looked through your working but shouldn't you neglect the "not divisible by 6" condition ? If a number is not divisible by 3, it won't be divisible by 6. That is redundant.

Edit : just noticed this was already pointed out by matt.
 
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I didn't neglect the redundance

Gokul43201 said:
Haven't looked through your working but shouldn't you neglect the "not divisible by 6" condition ? If a number is not divisible by 3, it won't be divisible by 6. That is redundant.

Edit : just noticed this was already pointed out by matt.
__________
I know that. I didn't neglect this redundance. The result is the same, not neglecting the redundance. I redo the calculus. But there is something wrong that I didn't find out yet.
 

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