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1. The problem statement, all variables and given/known data

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.

2. Relevant equations

The usual probability mumbo jumbo.

3. The attempt at a solution

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?

II.

1. The problem statement, all variables and given/known data

Out of the same set of 52, you draw 3 cards. What is the probability that at least one of those will be an ace?

2. Relevant equations

Same as above.

3. The attempt at a solution

Again, the solution (which I agree with this time) says that event will happen in (4 1)(48 2) + (4 2)(48 1) + (4 3) cases, which is logical. You figure out how many times you'll get on ace, how many times two aces and how many times three aces.

Where I'm lost here, however, is why can't you simply do (4 1)(51 2)? You secure one of those four aces with the first binomial, and then for the remaining two cards you can draw whichever out of the 51 that are left. But this leads to more events than described above and I just can't rationalize which additional events you get here that you shouldn't get.

Thanks in advance!

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# Homework Help: Two probability questions

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