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Why in Binomial distribution the formula starts with $n\choose k$ and not with something like $k!\over n!$? Isn't the order important? Or, it is important but due to the independence of the event?

gbox
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  • $k!n!\over n!$ would be the same as $k!$...as you sure that's what you meant to write? – man_in_green_shirt Jul 10 '15 at 13:26
  • Do you know what $\binom nk$ represents ? –  Jul 10 '15 at 13:34
  • @man_in_green_shirt right, I mean k fixed points in n permutations – gbox Jul 10 '15 at 13:34
  • @YvesDaoust yes, number of ways to choose k out of n with not repetitions and order does not matter – gbox Jul 10 '15 at 13:38
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    The binomial distribution is also order independent: you don't specify any order, do you ? –  Jul 10 '15 at 13:39
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    I recommend you to reconstruct binomial distribution trough an example forgetting absolutely all that you read in some book. By example count, graphically, the probability to get a 6 in n throws of a fair dice of six sides. You only need to know that probability is $\frac{\text{number of cases that I want to know it probability}}{\text{all cases}}$ – Masacroso Jul 10 '15 at 15:55

2 Answers2

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From the relevant Wikipedia page:

The binomial distribution is frequently used to model the number of successes in a sample of size $n$ drawn with replacement from a population of size $N$.

Note two things:

  • What is key is the number of successes. So if with four tries you have $(S,F,F,S)$ or $(F,S,F,S)$ is irrelevant, in both cases you have two successes.

  • Since one draws with replacement each try is independent from what happened before. Thus $(S,F,F,S)$ and $(F,S,F,S)$ certainly have the same probability.

And, $\binom{n}{k}$ is the number of ways in which you can have $k$ successes among $n$ tries.

If you do not draw with replacement, but without replacement, then the second point is no longer true and you get a different distribution, called hyper-geometric distribution.

quid
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    No. It is $\binom{2}{1}p (1-p)$ Generally it is $\binom{n}{k}p^k (1-p)^{n-k}$ where $n$ number of tries, $k$ number of successes, and $p$ probability of success. – quid Jul 10 '15 at 15:38
  • Sorry you respond very quick I will put it that way: If I want 1 "win" out of 6 tries, it turn out that it does matter in which try I won (1,0,0,0,0,0) or (0,1,0,0,0,0) etc. doesn't it mean that there is importance to the order? (unlike $n\choose k$ that does not has importance to order) – gbox Jul 10 '15 at 15:46
  • If you want to know what the probability is that you win exactly once, then it is $\binom{6}{1} p^1 (1-p)^5$. However, if you want the probability that you win in the third and only in the third try then the propability is $p^1 (1-p)^5$. The latter is not a binomial distirbution though. – quid Jul 10 '15 at 16:43
  • so at the first case there is importance to order, so (S,F,F,S) or (F,S,S,F) are 2 cases – gbox Jul 11 '15 at 17:55
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    Yes, they are two case, and you want to take both into account, together with all the other cases with two S and two F which are exactly $\binom{4}{2}$. The binomial coefficient $n$ over $k$ is there precisely to count the number of cases with $k$ wins in $n$ tries. You want to take all of them together, so you need to know how many there are. – quid Jul 11 '15 at 17:58
  • I am confused, doesn't it contradicts the first note you wrote in the answer, really trying to understand – gbox Jul 11 '15 at 18:10
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    Not really. You have the following possibilities with two success SSFF, SFSF, SFFS, FSSF, FSFS, FFSS. These are all different, but you only care about the number of successes. So you do not care which one of those happened. The prop for one particular of them to happen is $p^2(1-p)^2$ so you have $6 p^2(1-p)^2$ for one of the $6$ to happen. And the $6$ is just the $4$ over $2$. – quid Jul 11 '15 at 18:16
  • Really sorry but I am really trying to understand, on the one hand you say " You have the following possibilities with two success SSFF, SFSF, SFFS, FSSF, FSFS, FFSS.$ \textbf{These are all different}$" and then "So you do $\textbf{not care which one of those happened}$." and then you multiple it by 6 which is 4 over 2 – gbox Jul 11 '15 at 18:29
  • If you have a class with 20 children, and 13 are girls and 7 are boys and you select one at random (same probaility) then all the children are different Kate, Paul, Ali, Fan, Piotr, Cecile etc. and for one particular the prob is 1/20, but if you only care if you select a girl then you multiply the 1/20 by 13 the number of ways you can select a girl. – quid Jul 11 '15 at 19:05
  • Or in other words there is importance in which "attempt" the success was? – gbox Jul 11 '15 at 19:11
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    Let me give you still another example. You toss a coin three times, the possibilties are TTT, TTH, THT, HTT, THH, HTH, HHT, HHH. these are all posibilties. Each event has a prob of 1/8. Now it depends what you want to know. If you want to know the prob that you have H exactly once it is 3/8 as there are three different ways to get this. Precisely since you do not care which of the three it was, you have to multiply by the number of possibilties. By contrast the prob for HTT is 1/8. – quid Jul 11 '15 at 19:17
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$$\frac{n!}{n_1!n_2!},\;\; n_1+ n_2=n,$$

counts the permutations of the sequence $$\underbrace{p_1...p_1}_{n_1}\underbrace{p_2...p_2}_{n_2}.$$

Coincidentally, $$\frac{n!}{n_1!n_2!}=\frac{n!}{k!(n-k)!}={n \choose k}, \;\; n_1=k=\text{number of successes}.$$

Therefore, ${n \choose k}$, disguised as counting combinations, actually counts permutations.

James Pak
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