# Basic probability question **URL:** \ **Category:** Factual Questions\ **Created:** [February 16, 2005, 1:36pm UTC](https://boards.straightdope.com/t/basic-probability-question/290344 "2005-02-16T13:36:21Z")\ **Posts on this page:** 9\ **Page:** 1
**Author:** ![VunderBob](https://avatars.discourse-cdn.com/v4/letter/v/839c29/32.png) [@VunderBob](https://boards.straightdope.com/u/VunderBob)\ **Post date:** [February 16, 2005, 1:36pm UTC](https://boards.straightdope.com/t/basic-probability-question/290344/1 "2005-02-16T13:36:21Z")
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? ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [February 16, 2005, 1:52pm UTC](https://boards.straightdope.com/t/basic-probability-question/290344/2 "2005-02-16T13:52:21Z")
> [@VunderBob](#): > > 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. ---
**Author:** ![mks57](https://avatars.discourse-cdn.com/v4/letter/m/b9bd4f/32.png) [@mks57](https://boards.straightdope.com/u/mks57)\ **Post date:** [February 16, 2005, 1:55pm UTC](https://boards.straightdope.com/t/basic-probability-question/290344/3 "2005-02-16T13:55:24Z")
I’d sum the probabilities of the individual events to get the probability of any one, or more, of them occurring. So P = 0.3. ---
**Author:** ![Quercus](https://avatars.discourse-cdn.com/v4/letter/q/7ab992/32.png) [@Quercus](https://boards.straightdope.com/u/Quercus)\ **Post date:** [February 16, 2005, 2:20pm UTC](https://boards.straightdope.com/t/basic-probability-question/290344/4 "2005-02-16T14:20:04Z")
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 0.1_0.9_0.9 = 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) ---
**Author:** ![VunderBob](https://avatars.discourse-cdn.com/v4/letter/v/839c29/32.png) [@VunderBob](https://boards.straightdope.com/u/VunderBob)\ **Post date:** [February 16, 2005, 2:23pm UTC](https://boards.straightdope.com/t/basic-probability-question/290344/5 "2005-02-16T14:23:25Z")
> [@chrisk](#): > > Hope that helps. 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. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [February 16, 2005, 2:24pm UTC](https://boards.straightdope.com/t/basic-probability-question/290344/6 "2005-02-16T14:24:44Z")
> [@mks57](#): > > I’d sum the probabilities of the individual events to get the probability of any one, or more, of them occurring. So P = 0.3. 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%. ---
**Author:** ![mks57](https://avatars.discourse-cdn.com/v4/letter/m/b9bd4f/32.png) [@mks57](https://boards.straightdope.com/u/mks57)\ **Post date:** [February 16, 2005, 2:51pm UTC](https://boards.straightdope.com/t/basic-probability-question/290344/7 "2005-02-16T14:51:53Z")
I found this article from “The Math Forum @ Drexel” to be a useful explanation of when to add and when to multiply. [http://mathforum.org/library/drmath/view/56503.html](http://mathforum.org/library/drmath/view/56503.html) Somehow I got used to adding probabilities for reliability calculations without realizing that the answer was not correct, just “close enough”. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [February 16, 2005, 4:42pm UTC](https://boards.straightdope.com/t/basic-probability-question/290344/8 "2005-02-16T16:42:03Z")
The OP is asking for the binomial distribution. 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](http://mathworld.wolfram.com/BinomialCoefficient.html). ---
**Author:** ![Washoe](https://avatars.discourse-cdn.com/v4/letter/w/ecae2f/32.png) [@Washoe](https://boards.straightdope.com/u/Washoe)\ **Post date:** [February 16, 2005, 10:35pm UTC](https://boards.straightdope.com/t/basic-probability-question/290344/9 "2005-02-16T22:35:09Z")
Yes, ultrafilter is right—it’s a binomial probability distribution, and the equation is the right one. The probability of the event occuring is .243