# What is the probability of drawing at least one of each of these cards?

Say you have a deck of cards that contains $n$ total cards.

$x$ amount of these are Card A

$y$ amount of these are Card B

$z$ amount of these are Card C

By drawing $m$ cards at random from this deck, what is the probability of drawing at least one of each of these cards?

• Do we have $x+y+z=n$? – André Nicolas Aug 24 '15 at 19:33
• no, there would be cards that are not A, B, or C – Banbadle Aug 24 '15 at 20:16
• Are the types disjoint? Or are there cards of type A that are also type B, or similar combinations? (Eg: Type A is red cards, type B is Royal cards, Type C is $\{Q\heartsuit, A\spadesuit, 9\diamondsuit, 2\clubsuit\}$). That is, are there overlaps? – Graham Kemp Aug 24 '15 at 22:09
• Cards A B and C are different cards, no overlaps – Banbadle Aug 24 '15 at 22:24

Let $A$ denote the event that at least one of $A$ is drawn.

Let $B$ denote the event that at least one of $B$ is drawn.

Let $C$ denote the event that at least one of $C$ is drawn.

To be found is $P(A\cap B\cap C)$ and we have the equality:$$P(A\cap B\cap C)=1-P(A^c\cup B^c\cup C^c)$$

To find $P(A^c\cup B^c\cup C^c)$ make use of inclusion/exclusion.

I assume we are drawing without replacement. If necessary, put ID numbers on the objects to make them distinct. There are $\binom{n}{m}$ equally likely ways to choose $m$ cards from the $n$ available. Now we count the bad choices, in which one or more of the types A, B, or C is missing. By dividing by $\binom{n}{m}$ we can then find the probability of a bad choice, and hence of a good choice. We use the method of Inclusion/Exclusion.

There are $\binom{n-x}{m}$ ways to leave out Type A, with similar expressions for B and C.

But if we find the sum $$\binom{n-x}{m}+\binom{n-y}{m}+\binom{n-z}{m}\tag{1}$$ we will have double-counted the choices in which A and B are missing, also the choices where B and C are missing, also the ones where C and A are missing. So to count the bads, from the sum (1) we must subtract $$\binom{n-x-y}{m}+\binom{n-y-z}{m}+\binom{n-z-x}{m}\tag{2}$$

Nowever, we have subtracted too much, for we have subtracted one too many times the choices in which all of Type A, B, C are missing. There are $\binom{n-x-y-z}{m}$ of these.

Note that we are using the convention that if $k\lt l$ then $\binom{k}{l}=0$.