0
$\begingroup$

Given $n$ mutually independent Bernoulli trials each with success probability of $1/n^2$, what is the expected number of successes?

I think it should be $n/n^2$, but not sure. Can anyone help me with it?

$\endgroup$
1
  • $\begingroup$ you are right :) expected value of a Binomial distribution is n*p where n is the number of trials and p the probability of success. and n Bernoulli trials for a Binomial distribution $\endgroup$ – Albanian_EAGLE Dec 13 '13 at 3:52
0
$\begingroup$

You are correct. Here's a derivation of the result.

Let $X$ be the random variable that takes only values $0$ or $1$ and $$ P(X = 1) = \frac{1}{n^2}. $$

The variable that counts the total number of successes in $n$ trials is the sum $$ Y = X_1 + \cdots + X_n, $$ where each $X_i = X$. Therefore, $$ \begin{align} E(Y) &= E(X_1 + \cdots + X_n) \\ &= E(X_1) + \cdots + E(X_n) \\ &= \frac{1}{n^2} + \cdots + \frac{1}{n^2} \\ &= \frac{n}{n^2}. \end{align} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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