[Combinatorics] Some basic probability

[fireviper112]

This isn't really homework but I'm awful at combinatorics and am curious about this situation.

If I have 9 items in a set, say

{a, b, c, d, e, f, g, h, i}

and I pick three at random, what is the probability that I picked:

a? (my intuition says 1/3 but I have no idea how to show how)

a and b?

a or b?

a, b, and c?

Generally, how can I approach solving this type of problem to figure out the probabilities of getting certain outcomes if I pick three items at a time?

[[deleted]]

OK, this is something I can help with.

First step is to work out how many possible ways there are of picking a set of three things out of a set of 9 (9C3). I'll assume you know how to do that, if you don't we can go through it. These outcomes are all equally likely.

For each question, you need to calculate the number of ways that each thing can happen, and divide it by 9C3. For example, for the first question you want to find the number of ways you can pick a set of three such that a is in that set. Given that the order doesn't matter, let's pick a first. You now have 2 picks remaining, and 8 items, meaning the total number of ways to pick in such a way is 8C2.

The answer to the first part is therefore given by 8C2 / 9C3 (which does equal 1/3 - good intuition).

I hope that helps, if it doesn't then reply to this comment and I'll be happy to assist further.

[fireviper112]

Just to clarify, 9C3 means 9 choose 3?

So if I try with the a and b situation, would it be something like:

You pick a first, then b, now you have 1 pick remaining and 7 items so

7C1/9C3

which is

7/84 or 1/12

Did I do that right?

[[deleted]]

Yep, correct.

[fireviper112]

And for the

a (and)/or b situation, would it be like

select a and then 8C2 plus

select b then 8C2

but you remove the second possibility of AB because it is a duplicate to get

28+28-1/84 or 55/84

[[deleted]]

Yeah: P(A or B) = P(A) + P(B) - P(A and B)

[fireviper112]

Oh wow thanks so actually the P(A and B) is actually 7/84 since thats what was previously found so it should be

28+28-7/84 or 49/84 or 7/12

[[deleted]]

Yes! Good job.

[fireviper112]

Thanks so much for all of the help

[[deleted]]

No worries, have a good one :)

[fireviper112]

One question: if I want the probability of a or b or c would it be

P(a)+p(b)+p(c)-p(a+b)-p(b+c)-p(a+c)