I have a deck of 60 cards, all of which can be assigned to one of three categories (A, B, C). There are 30 cards in category A; 22 in category B; and the remaining 8 in category C. I intend to draw 7 cards from the deck. What I want to know is the probability of drawing at least one card from both categories B and C. Note that the order in which cards are drawn does not matter and they are not replaced once drawn.
I tried first figuring out the total number of possible outcomes by considering the different kinds of sets I can get by drawing 7 cards. These are the following:
AAAAAAA $\hspace{3mm}$ - $\hspace{3mm}$ 3
AAAAAAB $\hspace{3mm}$ - $\hspace{3mm}$ 6
AAAAABB $\hspace{3mm}$ - $\hspace{3mm}$ 6
AAAABBB $\hspace{3mm}$ - $\hspace{3mm}$ 6
AAAAABC $\hspace{3mm}$ - $\hspace{3mm}$ 3
AAAABBC $\hspace{3mm}$ - $\hspace{3mm}$ 6
AAABBBC $\hspace{3mm}$ - $\hspace{3mm}$ 3
AAABBCC $\hspace{3mm}$ - $\hspace{3mm}$ 3
The first kind of set listed above is just the set with all 7 cards of the same category. Though I only wrote all A's, it is understood that it can also be all B's or all C's (hence the number 3 to the right to indicate that there are 3 distinct sets of this type). So on and so forth.
Therefore, there are 36 distinct sets in total. By inspection, it can be seen that 15 of them contain at least one card of each category. On the other hand, 18 of them contain cards from two of the three categories, but only 6 of these correspond to having cards from categories B and C. Thus, 21 out of the 36 possible outcomes are the ones I desire. Nonetheless, the probability is not simply 21/36 because of the following reasons:
- The probability of drawing a card from category A is not the same as drawing a card from category B or C.
- Cards are not replaced after being drawn.
I have tried to look for well-known probability distributions, such as the hypergeometric distribution and the multinomial distribution, to see if they'd be more helpful. But frankly, I'm not very well versed in the theory behind them. So I was hoping if anyone could help me figure this problem out.
You could take it as a counting/combinatorics problem.
Probability that you don't have at least one from B and at least one from C is
$P_{\hbox{no}} = \frac{N(\hbox{All from A}) + N(\hbox{One or more from B, the rest from A}) + N(\hbox{One or more from C, the rest from A})}{\hbox{All combinations}}$
Then your desired probability is
$1-P_{\hbox{no}}$
So now you just need to count the sets "All from A", "One or more from B, the rest from A" and "One or more from C, the rest from A".
For "All from A", it is clearly $\binom{30}{7}$. For "One or more from B, the rest from A", you'll need to do a summation from 1 to 7, where the index of summation represents the number from B. Something like
$\sum_{k=1}^7 \binom{22}{k}\binom{30}{7-k}$
And likewise for "One or more from C, the rest from A"