rapsjade wrote:
Any number that is a multiple of both 8 and 2 is a multiple of 8. Because of the repetitive pattern of multiple of 8 among numbers, it should not make a difference whether we take 1000 as a sample or 10. There is one multiple of 8 between 1 and 10. so the probability is 1/8
Be careful - the part in
green isn't true.
If the question were about the first 10 positive integers, then the probability would be 1/10, since only 1 of the first ten positive integers is divisible by 8 (that integer being 8)
The integers from 1 to 1000 inclusive have a nice feature:
1, 2, 3, 4, 5, 6, 7,
8,
___ 9, 10, 11, 12, 13, 14, 15,
16,
___ 17, 18, 19, 20, 21, 22, 23,
24,
___ 25, 26, ...
As we can see, in every batch of 8 integers, exactly 1 is divisible by 8.
This patter continues all the way to 1000.
We have: ...
___985, 986, 987, 988, 989, 990, 991,
992 ___ 993, 994, 995, 996, 997, 998, 999,
1000So, all of our batches have exactly 8 integers.
As such, we can conclude that the probability is 1/8 that a selected number is divisible by 8
However, in the numbers from 1 to 10 inclusive, we have: 1, 2, 3, 4, 5, 6, 7,
8,
___ 9, 10
Here each batch does not have exactly 8 integers
So, we cannot conclude that the probability is 1/8 that a selected number is divisible by 8
Cheers,
Brent
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