Probability question (addition theorem of probability)

[chrome nimbus]

If a fair 15-sided dice is rolled, then is the probability that the roll is an odd number or prime number or both?
Can you please explain/solve it for me in detail

[clear edge]

there's nothing to solve

there's 15 cases, it would take 40 seconds

[shell umbra]

i mean theres 9 numbers on the dice that are either odd or prime

so its 9/15

(those numbers being 1,2,3,5,7,9,11,13,15)

[gray jay]

In mathematical formulas:
When X equals the number the dice gives, then the possible results are X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
You search the probability for X = {1, 2, 3, 5, 7, 9, 11, 13, 15} wich is equal to the sum of the probability for X = 1 and X = 2 and so on for each number in the brakets above.
This gives you P( X = {1, 2, 3, 5, 7, 9, 11, 13, 15} ) = 1/15 + 1/15 + 1/15 + 1/15 + 1/15 + 1/15 + 1/15 + 1/15 + 1/15 = 9 * 1/15 = 9/15

[chrome nimbus]

There is a prime number also and both how to solve it

[gray jay]

Forget what I wrote first. I oversaw, that theres the prime number condition.

[shell umbra]

we just look at all the numbers from 1 to 15 that are odd or prime

count them up, and divide that sum by 15

considering theres an equal probability of each thing to occur

[chrome nimbus]

I am solving it like this -
The odd number between 1-15 are 1,3,5,7,9,11,13,15
Prime numbers are 3,5,7,11,13
And both will be 1,3,5,7,9,11,13,15
By addition law a union b union c - p(a) +p(b) +p(c) -p(a inter b)-p(b intersect -c ) -p(a intersect c) + p(a union b union c)
It will be - 8/15+5/15+5/15-5/15-5/15-515+5/15 = 11/15

[shell umbra]

2 is also prime

but youre also overcomplicating it

it really is just a matter of adding up the individual probabilities of each value you want to get

you have 9 of those values

each with odds of 1/15

so 9 * 1/15 = 9/15

[chrome nimbus]

Ok