Choosing numbers without replacement What  is the probability of choosing 4 numbers from a bag of numbers from 1-10 without replacement, so that the smallest number choosen is 4?
I was thinking $\left(\frac{7}{10}\cdot\frac{6}{9}\cdot\frac{5}{8}\cdot\frac{4}{7}\right)$. 
Is it correct or am I wrong?
Thank you in advance!

After $4$ is fixed, the remaining three numbers have to be from $5$ to $10$ inclusive. So the number of draws whose smallest number is $4$ is the number of draws from $5$ to $10$, or $\displaystyle\binom{6}{3}=20$. Then the desired probability follows as $\frac{20}{\displaystyle\binom{10}{4}}=\frac{20}{210}=\frac2{21}$.
