Wikipedia:Reference desk/Archives/Mathematics/2008 December 18

= December 18 =

Simple probability
If an event has a probability p of producing a certain result, and n events are observed, what is the proabability that the result happens f times. This seems really easy, but I can't get it! I can work out the probability that it happens at least once, but I don't think that's relevent. Thanks. —Preceding unsigned comment added by 70.52.47.6 (talk) 04:32, 18 December 2008 (UTC)


 * A few hints: (1) In a sequence of $$n$$ events, what is the probability that the result in question is observed in exactly the first $$f$$ events, but not the other $$n-f$$? (2) Consider, more generally, any particular sequence of $$n$$ events in which the result is observed $$f$$ times. What is the probability that that particular sequence of $$n$$ events is observed? (3) In how many ways can you have exactly $$f$$ occurrences of the result distributed among $$n$$ events? --173.49.12.32 (talk) 05:23, 18 December 2008 (UTC)


 * Okay, unless I misunderstood you're hints/questions, the answers are 1: p^f, 2:(f!(n-f)!)/n!, 3: n!/(f!(n-f)!). Seeing as I'm not getting anywhere with this, I'll assume I misinterpreted your questions. Sorry. —Preceding unsigned comment added by 70.52.47.6 (talk) 05:41, 18 December 2008 (UTC)


 * Try making a probability tree - Mgm|(talk) 12:53, 18 December 2008 (UTC)
 * Also see binomial distribution. -- Jao (talk) 12:57, 18 December 2008 (UTC)


 * If there are f occurrences in n events, there must be n-f non-occurrences. The probability of an individual non-occurrence must be 1-p, so the first answer p^f is wrong, it should be p^f.(1-p)^(n-f) As that is the probability of one particular way of getting f occurrences in n events, you need to know how many different ways there are.→81.132.236.144 (talk) 13:56, 18 December 2008 (UTC)


 * Note that this is only true for n independent events. Imagine a coin that always alternates between head and tail. Each single toss has a probability of 50% to be heads, but a million-toss sequence will always contain *exactly* half a million heads. Taken literally, the original question is unanswerable. —Preceding unsigned comment added by 84.187.93.95 (talk) 01:39, 19 December 2008 (UTC)


 * You're mistaken. It's not ONLY if they're independent.  It's what happens IF they're independent, but not ONLY if.  For example, what if I toss a dime and quarter that are so entwined that if the dime comes up "tails" then so does the quarter, and that happens 1/4 of the time, and if the quarter comes up "heads" then so does the dime, and that happens 1/4 of the time.  The other 1/2 of the time consists of the dime showing "heads" and the quarter showing "tails".  Thus the numbers of "heads" that appear is either 0, ,1, or 2, with respective probabilities 1/4, 1/2, ,1/4&mdash;the same probabilities as if they were independent. Michael Hardy (talk) 01:58, 19 December 2008 (UTC)
 * They are in fact independent. Your formulation of the problem does not change that, the criterion for independence is the conditional probability.
 * No, they are not. We have P(dime shows tails and quarter shows heads)=0!=1/16=P(dime shows tails)P(quarter shows heads). Algebraist 01:03, 20 December 2008 (UTC)

The way the word "event" is used above conflicts with standard usage in the field. If I throw a die, getting an outcome of 5 or more is an event; getting an outcome of 4 or less is the complementary event. Throwing the die is NOT an "event"; sometimes it might be called an "experiment" or a "trial".

At any rate, binomial distribution is the article that addresses the question posed here (assuming the trials are independent). Michael Hardy (talk) 01:52, 19 December 2008 (UTC)


 * why when talking about Probability everybody is so definitely certain to be right? :) PMajer (talk) —Preceding unsigned comment added by 84.221.198.219 (talk) 09:16, 21 December 2008 (UTC)