What is the probability of selecting five of the winning balls and one of the supplementary balls? So I'm just doing a bit of probability questions and wanted to make sure I got it right. 
I have $50$ balls numbered $1-50$, and we pick $6$ winning balls and $2$ supplementary without replacement. 
So the chance to get the $6$ winning balls would simply be: $$\frac{6}{50} \cdot \frac{5}{49} \cdot \frac{4}{48} \cdot \frac{3}{47} \cdot \frac{2}{46} \cdot \frac{1}{45} = \frac{1}{15890700}$$
and for $5$ winning balls it would be, same process as above: $$\frac{3}{1059380}$$
Now the part that confuses me is $5$ winning, $1$ supplementary. Would this be given by: $$\frac{6}{50} \cdot \frac{5}{49} \cdot \frac{4}{48} \cdot \frac{3}{47} \cdot \frac{2}{46} \cdot \frac{44}{45} \cdot \frac{2}{44} = \frac{1}{7945350}?$$
Can someone please check if I did this right, I have a sense that I did not, but not sure where I went wrong.
 A: I will do it another way. The Tax on the Poor Corporation chooses $6$ "main" numbers and $2$ supplementary numbers. There are $\binom{50}{6}$ ways to choose the main numbers. Presumably they are all equally likely. 
We find the probability that you get all $6$ main numbers. There is only $1$ hand that will do the job, so the probability is $\frac{1}{\binom{50}{6}}$.
Now we find the probability that your hand has exactly $5$ main numbers, and none of the supplementary numbers. There are $2$ supplementary numbers, so you must have $5$ of the $6$ main numbers, plus one of the $42$ useless numbers. Thus there are $42\binom{6}{5}$ hands that have $5$ main numbers and no supplementary. That yields probability $\frac{42\binom{6}{5}}{\binom{50}{6}}$.
Finally, we count the hands that have $5$ main numbers and $1$ supplementary. There are $\binom{6}{5}\cdot 2$ such hands, and we compute the probability as before.
A: 
So the chance to get one of the $6$ winning balls would simply be: $\frac{6}{50}*\frac{5}{49}*\frac{4}{48}*\frac{3}{47}*\frac{2}{46}*\frac{1}{45} = \frac{1}{15890700}$

No, that is the chance to pick all of the six winning balls when you pick six numbers.   Exactly one winning ball among six picks would be:
$$\frac{6}{50}\times\frac{44}{49}\times\frac{43}{48}\times\frac{42}{47}\times\frac{41}{46}\times\frac{40}{45}\times 6=\dfrac{77572}{189175}$$
(With the last factor accounting for ways to place the non-winning ball in the draw.)

and for $5$ winning balls it would be, same process as above: $\frac{3}{1059380}$

$$\frac{6}{50}\times\frac{5}{49}\times\frac{4}{48}\times\frac{3}{47}\times\frac{2}{46}\times\frac{44}{45}\times 6 = \dfrac{22}{1324225}$$

Now the part that confuses me is 5 winning, 1 supplement. Would this be given by: $\frac{6}{50}*\frac{5}{49}*\frac{4}{48}*\frac{3}{47}*\frac{2}{46}*\frac{44}{45}*\frac{2}{44} = \frac{1}{7945350}?$

$$\frac{6}{50}\times\frac{5}{49}\times\frac{4}{48}\times\frac{3}{47}\times\frac{2}{46}\times\frac{2}{45}\times 6 = \dfrac{1}{1324225}$$
Note: the previous answers assumed you were not concerned about the count of supplemental balls.   The probability of $5$ winning and $0$ supplemental balls among six picks would be $\tfrac {21}{1324225}=\tfrac 3{189175}$.
