I tried this ,
No of prime no's below $500$ is $95$
Excluding $2,3,5,7$, it is $91$
No's not divisible $2,3,4,5,7,21$, are
$11 \cdot (11,13,17,19,23,29,31,37,41,43)$ i.e $10$
$13 \cdot (13,17,19,23,29,31,37)$ i.e $7$
$17 \cdot (17,19,23,29)$ i.e $4$
$19 \cdot (19,23)$ i.e $2$
initially $1$ is also not divisible
total = $91+10+7+4+2+1= 115$
answer $500-115 = 385$
Is this correct or incorrect??
It's seems that you've made some small mistakes, as the number of non-divisible numbers is 115, but there are $499$ integers less than $500$. Nonetheless your method might fail because you are prone to missing some numbers, like product of three primes greater than $7$, which isn't the case here, as $11^3 > 500$
Anyway probably the best way to go here is the Inclusion - Exclusion Principle. Namely let $N(c_1,c_2,...,c_n)$ is the number of elements that satisfy the condition $c_1, c_2,..., c_n$. Now if $c_1$ is the condition "divisible by 2", $c_2$ is "divisible by 3" and so on we have:
$N(\overline{c_1},\overline{c_2},\overline{c_3},\overline{c_4}) = N - \sum_{i} N(c_i) + \sum_{i,j} N(c_i,c_j) - \sum_{i,j,k} N(c_i,c_j,c_k) + N(c_1,c_2,c_3,c_4)$
The result will be:
$$499 - \left[\frac{499}{2}\right] - \left[\frac{499}{3}\right] - \left[\frac{499}{5}\right] -\left[\frac{499}{7}\right] + \left[\frac{499}{6}\right] + \left[\frac{499}{10}\right] + \left[\frac{499}{14}\right] + \left[\frac{499}{15}\right] +\left[\frac{499}{21}\right] + \left[\frac{499}{35}\right] - \left[\frac{499}{30}\right] - \left[\frac{499}{42}\right] - \left[\frac{499}{70}\right] - \left[\frac{499}{105}\right] + \left[\frac{499}{210}\right] = 115$$