<s>[INST] Probability of selecting the winning numbers in a lottery I've been studying combinatorics for a while. I've solved a problem but I'm not sure if I'm right. I'll just copy-paste the problem here.

In a lottery, six distinct numbers are selected at random from the set $1, \ldots, 50$ and designated as winning numbers. A player selects six distinct numbers in advance, hoping to include as many winners as possible. Find the probability that the player selects exactly $k$ winning numbers, for each $k$ from $0$ to $6$.

Here's my solution.
The probability of selecting one winning number is $\frac{\binom{6}{1}\cdot\binom{49}{5}}{\binom{50}{6}}$. And the probabilities of selecting 2, 3, 4, 5 and 6 winning numbers are $\frac{\binom{6}{2}\cdot\binom{48}{4}}{\binom{50}{6}}$, $\frac{\binom{6}{3}\cdot\binom{47}{3}}{\binom{50}{6}}$, $\frac{\binom{6}{4}\cdot\binom{46}{2}}{\binom{50}{6}}$, $\frac{\binom{6}{5}\cdot\binom{45}{1}}{\binom{50}{6}}$, and $\frac{1}{\binom{50}{6}}$ respectively. So the probability of selecting 0 winning number is
$1 - \left( \frac{\binom{6}{1}\cdot\binom{49}{5}}{\binom{50}{6}} + \frac{\binom{6}{2}\cdot\binom{48}{4}}{\binom{50}{6}} + \frac{\binom{6}{3}\cdot\binom{47}{3}}{\binom{50}{6}} + \frac{\binom{6}{4}\cdot\binom{46}{2}}{\binom{50}{6}} + \frac{\binom{6}{5}\cdot\binom{45}{1}}{\binom{50}{6}} + \frac{1}{\binom{50}{6}}\right)$
Am I wrong?
 [/INST] Just so this one doesn't stay unanswered, as Kamster says the chance of one winning number is $\frac{\binom{6}{1}\cdot\binom{44}{5}}{\binom{50}{6}}$ because when you choose the non-winning numbers there are only $44$ of them to choose.  Once you adjust your other answers with this in mind you will be correct.
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