We pick random a number of the set $[1000]$. Which the probability of that number is divisible for 4 but not for 5 and for 7? We pick random a number of the set $[1000]$. Which the probability of that number is divisible for 4 but not for 5 and for 7?
My work: 
Let $S=$"Set of solutions" then $|S|=1000$
Let $E=$"Pick a number divisible for 4 but not for 5 and for 7" a event.
Consider $E^c$ then
$E^c=\{35,70,105,140,175,210,245,280,315,3500,385,420,455,490,525,560,595,630,665,700,735,770,805,840,875,910,945,980\}$
This implies
$|E^c|=28$
Then, $P(E^c)=\frac{28}{1000}$
In consequence,
$P(E)=1-P(E^c)=0.9$
is good this?
 A: No, it's not good.  
First of all, you have some numbers in $E^C$ that should not be in there.  For example, $140 \in E^C$, but $140$ is divisible by $4$
Second, $1-\frac{28}{1000}=0.972$ .. you can't really 'round' that down to $0.9$.
Third, I believe you misinterpret the question: While I admit the question is a little ambiguous, I believe you should be looking for numbers that are divisible by $4$, but not by $5$ and also not by $7$. Thus, every number divisible by $5$ or by $7$ is in $E^C$
A: There are $250$ numbers in $1,2,\ldots, 1000$ divisible by $4$.  Of these, the $5^{\text{th}}$, $10^{\text{th}}$, and so on are divisible by $5$, the $7^{\text{th}}$, $14^{\text{th}}$ ... are divisible by $7$.  Of the $250$ numbers, $\lfloor 250/5 \rfloor $ are divisible by $5$, $\lfloor 250/7 \rfloor $ are divisible by $7$, while $\lfloor 250/35 \rfloor $ are divisible by both.
$$
|E| = 250 - \lfloor 250/5 \rfloor - \lfloor 250/7 \rfloor + \lfloor 250/35 \rfloor = 172
$$
The probability is $172/1000$.
