An urn contains 10 balls numbered from 1 to 10. Three balls are drawn without replacement. 
An urn contains 10 balls numbered from 1 to 10. Three balls are drawn without replacement.  What is the probability that the drawn balls have numbers greater than 5?

I thought for the first ball, (6,7,8,9,10). The probability is $5/10$. We took one, and for the second it is $4/9$. And for the third one it is $3/8$. Then we multiply these, and it is $1/12$.
But I am not sure for this solution. I am confused.
 A: Your solution is correct.
Another approach is to note that there are $\binom{5}{3}$ ways to select three of the five balls which are larger than $5$ and $\binom{10}{3}$ ways to select three of the ten balls, giving
$$\Pr(\text{three balls larger than $5$}) = \frac{\dbinom{5}{3}}{\dbinom{10}{3}} = \frac{10}{120} = \frac{1}{12}$$
A: Your answer is correct. Another way to look at the problem (and to arrive at the same answer) is the following counting argument:
Since the order in which we draw the balls doesn't matter and we do not put the balls back we have $\binom{10}{3}$ possible outcomes (number of ways to choose three elements out of ten elements where the order is irrelevant). Out of these
$$\binom{10}{3} = 120$$
possible outcomes (which are all equally probable) we have exactly $\binom{5}{3} = 10$ outcomes which satisfy that all three balls have numbers greater than $5$ (because this gives the number of ways to choose three balls from the numbers $6$ to $10$). The quotient
$$\frac{\binom{5}{3}}{\binom{10}{3}}= \frac{1}{12}$$
gives us the answer.
