Probability of a natural number being divisible by 2, 3, or 5? I'm trying to calculate the probability of a natural number being divisible by 2, 3, or 5 and I feel as if I may have found the answer.  But I wanted to see if anyone sees anything wrong with my "work".  Thank you all for your time and help.
Let ~ signify 'n is divisible by':
P[~2 ∨ ~3] = P[~2] + P[~3] - P[~2 ∧ ~3] = 1/2 + 1/3 - 1/6 = 2/3
P[(~2 ∨ ~3) ∨ ~5] = P[~2 ∨ ~3] + P[~5] - P[(~2 ∨ ~3) ∧ ~5] = 2/3 + 1/5 - something
something = P[(~2 ∨ ~3) ∧ ~5] = P[(~2 ∧ ~5) ∨ (~3 ∧ ~5)] = 1/10 + 1/15 - 1/30 = 4/30 = 2/15 so
P[(~2 ∨ ~3) ∨ ~5] = 2/3 + 1/5 - 2/15 = 11/15
Are these calculations correct and am I even using probabilities and such correctly?
 A: You have to describe how you randomize a natural number.  It is not possible to have a discrete uniform distribution on $\mathbb{N}$.  If you are talking about the natural density of natural numbers divisible by $2$, $3$, or $5$, then the answer is $\frac{11}{15}$.  Alternatively, if you are talking about the discrete uniform distribution on $\mathbb{Z}/30\mathbb{Z}$, then your calculation is correct.
A: It is quite simple: $$P=1-\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac15\right)$$
A: Out of the $30$ congruences $\bmod 30$ there are $22$ that work. So I would say under any sensible distribution the probability should be $\frac{22}{30}=\frac{11}{15}\approx0.73$.
A: We want to find the probability that a natural number be divisible by either $2$, or $3$, or $5$. We note a few things first. All numbers are prime, so the probabilities $P(2), P(3), P(5)$ are independent (but not mutually exclusive). A chosen number may be divisible by both $3$ and $5$. But this scenario doesn't add complexity, since we require for the number to be divisible by at least one of them. We can use the standard method for calculating the probability of the union of 3 events, namely $P(2), P(3)$ and $P(5)$ which takes care of all the conditions: 
$$ P(2 \lor 3 \lor 5) = P(2) + P(3) + P(5) - P(2\land 2) - P(2 \land 5) - P(3 \land 5) + P(2 \land 3 \land 5) = 1/2 + 1/3 + 1/5 - 1/6 -1/10 - 1/15 + 1/30 = \ldots = 11/15 = 73.\overset{.}{3}\ \% $$
