Let $X=\{0,1,...,9\}$. A random 4-tuple, $(a_1,a_2,a_3,a_4)$ is selected from $X^4$. I am calculating the probability that the least digit is 3.
My attempted solution is: We need to select 3 and then three other digits larger than 3, which if they occur in that order, have probability
$$\frac{1}{10}\cdot\left(\frac{6}{10}\right)^3$$
However, since these digits may occur in any order, we should multiply by $4!$ in order to count all rearrangements.
Something about this seems too easy though--is this solution valid? If I were to calculate the probability that the 2nd largest digit is 3, would it have solution
$$\frac{1}{10}\cdot \frac{3}{10}\cdot\left(\frac{6}{10}\right)^2\cdot 4!$$
In this case, you have four types of favorable cases (according to the number of 3's in the vector) Favorable cases: $\binom{4}{1}6^3+\binom{4}{2}6^2+\binom{4}{3}6+\binom{4}{4}$
The total number of cases are $10^4$.
By making the divison you get a probability of approximately $0.1684$