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The $p$ (probability of success) parameter of binomial distribution $Bin(n,p)$ can be estimated by deriving it from the mean:

$$\mu = np \iff p = \frac{\mu}{n}$$

However, given example that r.v. follows $Bin(n,p)$ and in an experiment 2 out of n=10 binomial trials are successes (i.e. the modellable event is $P(\text{2 out of 10 are successes})$), then shouldn't $p$ be $$p=\frac{2}{10}$$

But is 2 the mean value of this trial?

mavavilj
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    Note: the binomial distribution models the number of successes, not the probability of success. – ocram Jan 20 '16 at 14:02
  • Surely $np$ is the expected value not the mean? – Björn Jan 20 '16 at 14:03
  • @ocram So even if $p=0.5$ for a single trial out of the 10, the trial of all 10 should have a different $p$, the one that corresponds to $P(\text{2 out of 10 are successes})$? – mavavilj Jan 20 '16 at 14:09
  • @Björn I've led to believe that generally $E(X)=\mu$ in binomial distributions. http://www.stat.yale.edu/Courses/1997-98/101/binom.htm (see "Mean and Variance ..."). – mavavilj Jan 20 '16 at 14:11
  • The wording in Wikipedia seems a bit off as well: "p is the discrete probability distribution of the number of successes in a sequence of n independent yes/no experiments, each of which yields success with probability p.". So is $p=0.5$ or something else? https://en.wikipedia.org/wiki/Binomial_distribution – mavavilj Jan 20 '16 at 14:42
  • Here it's said that "This provides an estimate of the parameter p, the proportion of individuals who support the candidate in the entire population." But also "The probability of "success" p is the same for each outcome." http://www.stat.yale.edu/Courses/1997-98/101/binom.htm – mavavilj Jan 20 '16 at 14:56
  • @mavavilj if each person of the population supports some candidate with the same probability $p$ then if we have $n$ individuals, than on average $np$ will support the candidate. – Tim Jan 20 '16 at 15:00

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It seems that you are misunderstanding what $p$ is in binomial distribution. We use $p$ to denote one of the parameters of binomial distribution. So $x$ may have binomial distribution with parameters $n$ and $p$

$$ x \sim \mathrm{Binom}(n,p)$$

and the probability of $X = x$ is

$$ f(x;n,p) = \Pr(X = x) = \binom n x p^x(1-p)^{n-x} $$

so $p \neq \Pr(X = x)$. As you said, $\mu = np$ and $p = \frac{\mu}{n}$, so if you get on average $2$ successes out of $10$ trials then $p$ is $2/10$.

I'll translate this to simple example. You have biased coin (no matter that such thing probably doesn't exist) with probability of obtaining heads in a single flip is $p$ and of tails is $1-p$. Single coin flip follows a Bernoulli distribution with parameter $p$ (so $p = 1/2$ is a fair coin). Now if you want to calculate probability of obtaining $x$ heads after flipping such coin $n$ times (i.e. $n$ Bernoulli trials), you use binomial distribution with parameters $n$, $p$ and on average you get $np$ heads. Each of the coin flips is independent of other and probability of heads in each of the flips is the same and equal to $p$.

Tim
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  • But shouldn't the success of a single trial be 0.5 if a single trial (a single out of the 10) can either succeed/fail. Then what's $p=\frac{2}{10}$? Or does it mean that $p=\frac{2}{10}$ is the probability of the event $P(\text{2 out of 10 is successes})$ and then "a single binomial trial" in this case means "a set of 10 trials". – mavavilj Jan 20 '16 at 14:29
  • @mavavilj Only if it is a fair coin, i.e. you are talking about Bernoulli distribution with $p=1/2$... – Tim Jan 20 '16 at 14:31
  • So what's $p$ in your answer? If "on average 2/10 successes", then it means that the mean is 2/10? And then $p=(2/10)/10=0.02$? – mavavilj Jan 20 '16 at 14:35