# What is the percentage chance of drawing at least 1 of 12 cards when drawing 3 of a deck of 44 cards?

What is the percentage chance of drawing at least 1 of 12 cards when drawing 3 of a deck of 44 cards?

I'm not sure how to calculate this. It's related to a game that a play, so it's not homework or anything.

I know that 44 choose 3 would be the number of combinations that could be drawn, but how do I narrow it down to just check if at least one of them are in a preset group of 12 of the 44 cards? (How many combinations of 44 choose 12 contain at least one of the three cards?)

If someone could explain the process of calculating this, that would be very helpful. Thanks.

• It would be easier to compute the probability that none are drawn. Then subtract this quantity from $1$ to get your desired probability. – David Mitra Jun 6 '15 at 18:27

$$P(\text{none}) = \frac{\binom{32}{3}}{\binom{44}{3}}$$
$$P(\text{at least one}) = 1 - P(\text{none})$$
• @mbomb007 Same idea. $P(\text{at least two}) = 1 - P(\text{none}) - P(\text{exactly one})$. – Barry Jun 6 '15 at 22:33