Really simple probability question

I'm having trouble calculating some probabilities for a simple card game I'm into

1) So say we have a deck of 13 cards total. You draw 3 cards so there are 10 cards left in the deck. Say there is 1 special card in the deck that we want to figure out the probability of getting it in the hand. How do we figure it out? Would it be:

$$\dbinom{1}{1} * \dbinom{12}{2} / \dbinom{13}{3}$$

1 choose 1 for the special card. 12 choose 2 for the other 2 cards which we don't care what they are. I was thinking I might have to multiply by 3! because there are 3! ways to order the 3 cards in the hand but that would make the probability greater than 1.

2) Now let's say in the deck there is that 1 special card, and also 3 copies of a different type of card that we want. How would we calculate the probability of forming a hand that contains the 1 special card AND 1 or more of any of the 3 copies of the 2nd type of card?

3) Now let's say we have a deck with 2 copies of card type A, and 2 copies of card type B. How would we calculate the probability of choosing 1 or more from type A AND 1 or more from type B assuming a 13 card deck where we draw 3 cards? (for example: 1 type A + 1 type B + 1 any other card, 1 type A + 2 type B, etc).

I remember doing math like this in high school and it being really basic but don't quite remember exactly how to solve them. Thanks!

• Your answer is on 1) is correct. Multiplying by $3!$ is okay if you consistently also multiply the denominator by $3!$. These acts cancel eachother. – drhab Jan 10 '15 at 18:35

Looks right.

"I was thinking I might have to multiply by 3! because there are 3! ways to order the 3 cards in the hand " --> No need. Order doesn't matter in this case.

"Now let's say in the deck there is that 1 special card, and also 3 copies of a different type of card that we want. How would we calculate the probability of forming a hand that contains the 1 special card AND 1 or more of any of the 3 copies of the 2nd type of card?"

1 special card, 1 of 2nd type

$\binom{1}{1} \binom{3}{1} \binom{9}{1} / same$

1 special card, 2 of 2nd type

$\binom{1}{1} \binom{3}{2} \binom{9}{0} / same$

1 special card, 3 of 2nd type

impossible since we draw 3

"Now let's say we have a deck with 2 copies of card type A, and 2 copies of card type B. How would we calculate the probability of choosing 1 or more from type A AND 1 or more from type B assuming a 13 card deck where we draw 3 cards? (for example: 1 type A + 1 type B + 1 any other card, 1 type A + 2 type B, etc)."

X type A, Y type B

$\binom{2}{X} \binom{2}{Y} \binom{10}{3-X-Y} / same$

where X = 1 or 2, Y = 1 or 2 and X + Y $\leq 3$

• Thanks for the response. When you answered the 2nd question for 1,2,3,2 or 3 special cards, does that also take into account having having the original 1 special card in addition to the 2nd type of card (just to clarify there's 1 special card which i'll call type X and 3 copies of a 2nd special card i'll call type Y. I was asking about having 1 copy of type X and simultaneously having 1 or 2 copies of type Y). Also for the 3rd question, how exactly does the combinatoric notation work when having multiple values in the bottom for C(10, 3 X Y )? – km13818 Jan 10 '15 at 19:21
• On Q2: No ( why not type A and type B like I did? :( ). Sorry I completely misread the question. Will edit. On Q3: Huh? That is 10 choose the number which is 3 minus X minus Y. If X = 1 and Y = 2, then the number is 0. – BCLC Jan 10 '15 at 19:42
• @km13818 edited – BCLC Jan 10 '15 at 19:46
• Oops i didnt see the minus signs lol. I got it now, thank you! – km13818 Jan 10 '15 at 21:27

Hint on 2)

$P(\text{the special AND}\geq1 \text{ copies of...})=P(\text{the special})-P(\text{the special AND 0 copies of...})$