Probability of a single number from a set $\{1,2,3...1000\}$ being a multiple of any two $7,11$ or $13$ I am asked to pick a single number uniformly at random from the set $\{1,2,3...1000\}$. What is the probability that it's a multiple of any two of the numbers $7,11$ or $13$.
It would be very helpful for me to visualize the solution by observing how one should approach this type of problem
 A: Hint : You can prove that that the number of integers below $n$ divisible by $k$ is equal to $\lfloor \frac n k\rfloor $. Then by using inclusion–exclusion principle you should be able to conclude.
A: If a number is a multiple of any two of the numbers $7$, $11$, or $13$, then it divides either $7*11 = 77$ (there are $\lfloor \frac{1000}{77} \rfloor = 12$ of them), or $11*13 = 143$ (there are $\lfloor \frac{1000}{143} \rfloor = 6$ of them) or $7*13 = 91$ (there are $\lfloor \frac{1000}{91} \rfloor = 10$ of them). One can also note that those three classes of numbers do not intersect, as $7*11*13 = 1001 > 1000$. So the total number of such numbers is $28$. And as our numbers are uniformly distributed, our probability is $\frac{28}{1000}$.
A: To be a multiple of any two numbers, it must be a multiple of the least common muliplte (LCM) of those two numbers. In this case, all three numbers are prime so the LCM of any two is simply the product of the two. Here we have $$7*11=77,\text{ }7*13=91, \text{ and }11*13=143$$
If you divide each $product%$ into $1000$, you will find the number of multiples of that product. 
Now, since $$\text{INT}(1000/77)=12,\text{ INT}(1000/91)=10,\text{ and INT}(1000/143)=6,$$
we have $12+10+6=28$ chances out of a thousand of drawing a multiple of any two of $7,11,\text{ or }13$. Subtract three if you do not count 1 $times$ as a multiple.
