# Thread: Probability (drawing cards out of a set of 52)

1. ## Probability (drawing cards out of a set of 52)

Hey guys, I'm new here, but already have a question

Basically, the task at hand is as follows. You have a set of 52 cards, out of which you randomly draw 8. What's the probability of an event that either a) three aces will be drawn or b) three kings will be drawn or c) three aces and three kings will be drawn.

This is from a high school textbook and the solution says the said events are going to happen in (4 3)(48 5) + (4 3)(48 5) + (4 3)(4 3)(44 2). I get the first two, but I don't get why there is a plus in front of the third. Shouldn't there be a minus? I mean, the event c) is already covered in both the events a) and b), so if anything you should subtract that event so that it isn't counted twice. If my thinking isn't correct, what am I missing?

Thanks in advance, everyone, I've been doing this problem for hours, but still can't figure out where I'm going wrong. Oh, and sorry for not using Latex, I'm not familiar with it. Those (4 3) etc. are supposed to represent binomial symbols, with 4 being "n" (at the top) and 3 being "r" (on the bottom).

2. does (4 3) means ${4 \choose 3}\; ?$

3. Originally Posted by Ryker
Hey guys, I'm new here, but already have a question

Basically, the task at hand is as follows. You have a set of 52 cards, out of which you randomly draw 8. What's the probability of an event that either a) three aces will be drawn or b) three kings will be drawn or c) three aces and three kings will be drawn.

This is from a high school textbook and the solution says the said events are going to happen in (4 3)(48 5) + (4 3)(48 5) + (4 3)(4 3)(44 2). I get the first two, but I don't get why there is a plus in front of the third. Shouldn't there be a minus? I mean, the event c) is already covered in both the events a) and b), so if anything you should subtract that event so that it isn't counted twice. If my thinking isn't correct, what am I missing?

Thanks in advance, everyone, I've been doing this problem for hours, but still can't figure out where I'm going wrong. Oh, and sorry for not using Latex, I'm not familiar with it. Those (4 3) etc. are supposed to represent binomial symbols, with 4 being "n" (at the top) and 3 being "r" (on the bottom).
I agree with you that the third term should be subtracted, not added. This is an example of the inclusion-exclusion principle.

If they want a plus they could do

$\displaystyle \binom{4}{3}\bigg[\binom{48}{5}-\binom{4}{3} \binom{44}{2} \bigg] + \binom{4}{3}\bigg[\binom{48}{5}-\binom{4}{3} \binom{44}{2} \bigg] + \binom{4}{3}\binom{4}{3}\binom{44}{2}$

4. Thanks, I'm really glad to see I wasn't mistaken, though I hate the fact that I lost quite a lot of time over this just because the damn solution wasn't correct.