# Random variable and trying to find $E(N)$

Given $N = \text{random variable that counts the fraction of trials that are successful trials} = 12$

$N = S/12$

$S = \text{number of successful trials}$

$E(N)= ?$

i don't know how to find $E(N)$ is there a specific formula? can someone help me out with this problem

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You can calculate the probability of a single experiment succeeding, right? Call that $p$, as it's the same for every other experiment. So imagine each experiment as a whole being a single coin flip with probability of success $p$. So you're looking for the expected number of successes after 12 flips. Do you know which distribution this follows? – Alex R. Mar 27 '13 at 18:13

It is useful to introduce some notation. Define random variables $X_1,X_2,\dots, X_n$ by $X_k=1$ if the $k$-th trial is successful, and $X_k=0$ if the trial is not successful.
Then $S=X_1+X_2+\cdots +X_n$: the sum of the $X_i% counts the number of successful trials. Finally, note that$N=\dfrac{S}{n}$:$N$is the proportion of successful trials. Then $$E(S)=E(X_1+X_2+\cdots +X_n)=E(X_1)+E(X_2)+\cdots +E(X_n).$$ We calculate$E(X_i)$. The probability that$X_i=1$is the probability of$2$heads, which is$\frac{1}{4}$. So$E(X_i)=(1/4)(1)+(3/4)(0)=1/4$. It follows that$E(S)=\dfrac{n}{4}$. Note that$E(N)=E\left(\dfrac{1}{n}S\right)=\dfrac{E(S)}{n}$. Thus$E(N)=\dfrac{1}{4}$. Remark: The answer is intuitively clear. Actually, the number of successes in$n$trials has binomial distribution with "$p$", the probability of success, equal to$1/4$. The proportion of successes should have expected value$\frac{1}{4}\$. We introduced the random variable machinery because it will be long run useful.