# Divisible of natural numbers

1. Sep 11, 2005

### Alexsandro

Could Someone help with this question of probability ?

What is the probability that a number of the set $\Omega$, 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 ?

Last edited: Sep 11, 2005
2. Sep 11, 2005

### Hurkyl

Staff Emeritus
Well, the first problem is to figure out just what you mean by "picking a natural number at random".

(P.S. there is no such thing as a uniform distribution on the naturals)

3. Sep 11, 2005

### cefarix

Answer: (1/4 - 1/12) + (1/5 - 1/15 - 1/20)

Explanation:
Half of the numbers will be divisible by two. Out of these, another half will be divisible by four and shall be excluded, hence we end up with 1/4. These numbers will be of the form 4n+2, and none of them will be divisible by 3. However, 1 out of 3 of these will be divisible by 6, so we subtract 1/3 * 1/4 = 1/12 from 1/4 to get those numbers divisible by 2 but not by 3,4, or 6.
One fifth of the numbers are divisible by five. Out of these, multiples of 15 will be divisible by 3, multiples of 20 will be divisible by 4, and multiples of 30 will be divisible by 6. However, half of the multiples of 15 are also multiples of 30, so we don't need to include a 1/30 term, as it is already covered by the 1/15 term. So the probability of a number being divisible by 5 but not by 3,4, or 6 is 1/5 - 1/15 - 1/20.
Adding the two probabilities together gives the final answer for a random natural number being divisible by either 2 or 5 but not by 3,4, or 6.

Last edited: Sep 11, 2005
4. Sep 11, 2005

### Alexsandro

Sorry, I forgot to say that this question refers to set $\Omega$ of the first 120 natural numbers {1,2,3, ... , 120}. I'd like to know what probability that a natural number this set $\Omega$ is not divisible by 3, 4 and 6 but is divisible by 2 or 5.

I thank if someone help me!

5. Sep 11, 2005

### Berko

Brute force doesn't hurt. If it is divisbile by 2 and 5, then it is divisible by 10. There are only 12 multiples of 10 in the first 120 numbers. Just count those which are not divisible by 3,4, or 6.

6. Sep 11, 2005

### Alexsandro

What is the probability that a number of the set $\Omega$, 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 ?

The divisible if for 2 OR 5, not 2 AND 5 !!!

7. Sep 11, 2005

### Berko

Ok, I stand corrected. My apologies on the error in reading.

Nevertheless, the brute force approach still works. Use the sieve of Eratothenes (write all numbers from 1 to 120 and cross them off as they violate your requirements).

Then divide the number of numbers that remain by 120.

8. Sep 12, 2005

### shmoe

This is too low- you've removed the numbers divisible by 5 and not divisible by 3 AND 4 twice. Similar problem for your end result, you need to also remove the numbers divisible by 2 AND 5, but not 3 and 4 as these were counted twice.

Alexsandro, what are you able to do on this problem? Post some work and it will be easier to offer some help. Can you answer some simpler questions like:

What is the probability a number is divisible by 2 and not divisible by 3?

What is the probability a number is divisible by 2 and not divisible by 3 and 4?

9. Sep 14, 2005

### Alexsandro

______________________
Ok Shmoe,
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?

10. Sep 15, 2005

### Mithal

If it is divisible by 2 you have 60 choices

1 , 2 , 3 ...... ,59 , 60 all multiply by 2

we remove 60/2 +60/3 -60/6=40

it remains 60-40 =20

if it divides 5 we have 1,2,3 ,........,24 all multiply by 5

we remove 24/3 +24/4-24/12=12
what remains 24-12=12
We have four common cases above
2*5=5*2
2*25=5*10
2*35=5*14
2*55=5*22
then
20+12-4=28
p=28/120=7/30

Last edited: Sep 15, 2005
11. Sep 15, 2005

### Alexsandro

doubt

_______________
I didn't understand the removings that you did. Please, could you explain it.

"we remove 60/2 +60/3 -60/6=40
it remains 60-40 =20 "

and

"we remove 24/3 +24/4-24/12=12
what remains 24-12=12
We have four common cases above
2*5=5*2
2*25=5*10
2*35=5*14
2*55=5*22
then
20+12-4=28
p=28/120=7/30"

thanks

12. Sep 15, 2005

### Mithal

If we have

(1,2 ,3 ,... ,59,60 all multiplied by 2

and we want to find the number the divides 2 or 3)

that is equivalent to finding the number that divides 4 or 6 in

the numbers 2,4,6,8 .......,118,120

So if we have

1,2,3,.... ,59,60

and we want to remove the numbers that divides 2 or 3

First we remove the numbers that divides 2

which is 2,4 ,6 ,... ,58,60

Which is equal to 60/2=30

Now the numbers that divides 3

3,6,9 ,......, 57,60

which is equal to 60/3=20

Now we removed twice the number that divides both 2 and 3 which is

multiple of 6

6,12,18,.... ,60

or equal to 60/6=10

Now we removed 30+20 -10=40

it remains 60-40=20

The same thing for 5

Now we try to figure the numbers that are common on both counting of 2

and 5

we have among the set (1,2,....59,60) that divides 5 are

5,10,15,20,25,30,35,40,45,50,55,60

We exclude the numbers the divides 2 or 3 . We have then the four cases as

follow

5,25,35,55

finally the number that satisfies the conditions of the problem=20+12-4=28

p=28/120