[Math] An integer is chosen at random from the first 1000 positive integers. Probability that is a multiple of both 6 and 8

Question

An integer is chosen at random from the first one thousand positive integers. Find the probability that the integer chosen is:

a. a multiple of 6
b. a multiple of both 6 and 8

I've done a, $1000/6=166$ multiples of 6 between 1 and 1000
$166/1000=0.166$

For b, I presume it would be something like:

$\dfrac{1000}{6}+\dfrac{1000}{8}-x$ where $x=\dfrac{1000}{\text{multiples of 6 and 8}}$

But how do you work out the number of integers that are a multiple of both 6 and 8?

Answer 1

Hint: For a number to be a multiple of both $6$ and $8$, it has to be a multiple of $\operatorname {LCM}(6,8)=24$