If I had my discrete mathematics text handy, I could get an answer to this in about 30 seconds. The book, however, is buried in a storage unit, and I’m too lazy to spend an afternoon digging through boxes to find it.
If you have an individual event with x probability of occuring, in a setting of N independent events, what is the probability of a subset y occuring?
If you need a face to the problem, say there are 3 groundhog holes, and the odds of seeing a groundhog sticking his head out at any time is 1/10. What are the odds of seeing 1 groundhog head?
Okay… based on the theoretical description, I’m assuming that probability of 1/10 is the chances of getting groundhog at any one specific hole at a time, and I’m guessing that you’re interested in the odds of getting exactly one groundhog out of the three holes taken together. So, let’s build it up a bit at a time.
For the first two holes, you have 1/10 * 1/10 or 1/100 of seeing two groundhogs at the same time, 9/10 * 9/10 or 81/100 of seeing no groundhogs at all, and (1/10 * 9/10) + (9/10 * 1/10) or 18/100 of seeing one groundhog total.
To factor in the third hole, the chance of seeing exactly one groundhog would be (18/100 * 9/10) + (81/100 * 1/10) …
0.162 + 0.081…
0.243 or 243/1000
If you’re interested in the odds of getting one OR MORE groundhogs, btw, that can be figured out first from the odds of seeing none at all (9/10 * 9/10 * 9/10 or 729/1000) and then subtracting that from 1 to get 271/1000
Hope that helps. If you’re unclear on some of the principles involved, just ask about whatever confuses you.
Upon preview, Chrisk got it as well, but here’s my answer with slightly different route.
I’m assuming you mean exactly one groundhog.
The chance of groundhog in hole 1 and none in the others is 0.10.90.9 = 0.081
And you can have a groundhog in hole 1,2, or 3, so multiply by 3
3*0.081 = 0.243
I think the general formula for k out of n occurences with chance p each is
p^k*(1-p)^(n-k)*(# of ways to choose k out of n)
It did. I was rather good at the probability stuff taking the class, but I don’t use it on the job, and I’ve slept a few times because the class was 4 years ago.
mks57, summing works as an approximation, but the exact answer is what I was looking for. 0.3 is close to the exact answer of 0.271. The difference in the two numbers will be multiple versions of the same combinatorial case.
But that wouldn’t be correct, as you can see from an extension of VunderBob’s thought experiment: Suppose there are eleven groundhog holes. Extending your methodology, the overall probability of seeing a groundhog is 110%.
If X is binomial(n, p), the probability that X = k is given by [sub]n[/sub]C[/sub]k[/sub]p[sup]n[/sup](1 - p)[sup]n - k[/sup], where [sub]n[/sub]C[/sub]k[/sub] is a binomial coefficient.