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There is a deck of $30$ cards, each card labeled with a unique number from $1$ to $30$. If you draw a hand of $10$ cards, what is the probability that both the $1$ and $2$ cards are in your hand after drawing this?

I just want to check my work is correct:

There is only one way to choose each card since they are unique in the deck, so $\binom{2}{2}$ ways to choose the $1$ and $2$ cards. Then there are $\binom{28}{8}$ ways to choose the rest of the $8$ cards in your hand. So we multiply these two and divide them by the total number of possibilities of $10$ card hands: $\frac{\binom{2}{2} \cdot \binom{28}{8}}{\binom{30}{10}} = 10.35\%$. Is this right? Thanks

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    $\begingroup$ It looks right to me $\endgroup$ – Henry Sep 3 '17 at 8:05
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Yes, you are correct. Note that since the cards are all distinct we can put it also in this way. The probability that the $1$ is in your hand of 10 cards is $10/30$. Given that, the probability that the $2$ is among the remaining 9 cards is $9/29$. Hence the probability that both the 1 and 2 are in your hand is $$\frac{10} {30}\cdot\frac{9}{29}=\frac{3}{29}$$ which coincides with your result because $${\binom{30}{10}}=\frac{30\cdot 29}{10\cdot 9}\cdot {\binom{28}{8}}.$$

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I'm getting the same (28C8)/(30C10) but it's coming to 10.34.

However, IMO 2C2 not needed to be considered as I'm considering '1' & '2' already in my hand while considering 28C8.

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