# Bernoulli trials

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?

• 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 – Albanian_EAGLE Dec 13 '13 at 3:52

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}