What are the chances of picking any 9 out of 12

Question

Math help

Answer 1

you can approach this as a Combination problem because the order in which you pick them out does not matter

C(12, 9) = 12! / 9!(12 - 9)! = 220

meaning that there are 220 different ways that you could pick 9 items out of 12

Answer 2

Hi Paul.

 

For questions like these, one way of answering them is to think about it 1 choice at a time.

 

When you start out, you have 12 options to choose from, so the odds of choosing any particular item is 1/12.

 

Once you've made that selection, there are only 11 options left. So the probability of picking any particular item from the remaining 11 options is 1/11.

 

Another thing to keep in mind is that to find the probability of two independent events occuring together, you multiply the individual probabilities. So what are the odds that you choose any 2 out of 12?

 

P = (1/12) (1/11) = .007576... or 0.7576%

 

The pattern continues for every choice, where after each one, there is 1 less option available. So, since you end up choosing 9 items total, the probability of choosing a specific set of 9 (where the order they were selected matters) is

 

P(specific 9 out of 12) = (1/12) (1/11) (1/10) (1/9) (1/8) (1/7) (1/6) (1/5) (1/4) = 0.000000012526 or 0.0000012526%

 

Now, what if the order doesn't matter? Once way to think about it to ask, how many different ways could they have been ordered? For example, if choosing A then B is no different than choosing B then A, I have to add those probabilities together. P(A and B) = P(A then B) + P(B then A) = 2 P(A then B).

 

The number 2 comes from the fact that there are 2 ways to arrange 2 items:

 

A,B

B,A

 

What if you have 3?

 

A,B,C

A,C,B

B,A,C

B,C,A

C,A,B

C,B,A

 

Then there are 6 ways to choose the same set ABC. Now, this counting gets pretty tricky as the numbers get bigger, which is why we have something called a factorial, represented by an exclamation point. N! = N (N-1) (N-2) ... (1) is the number of ways to arrange N elements. So in the example of 3, we counted 6 ways to arrange them, and 3! = (3) (2) (1) = 6 as well. 

 

So, back to the question of choosing any 9 out of 12. We already found that 

 

P(specific 9 out of 12) =  0.000000012526

 

So, if order doesn't matter, we should have P(9 out of 12) = (9!) (0.000000012526) = .004545 or 0.4545%

 

Now, once you are familiar with these ideas and where they come from, there are also methods involving something called n choose k, but this post is already long enough, so I'll stop there.

 

Hope this helps!