Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

a busy cat

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

share|cite|improve this question
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
up vote 2 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.