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?
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$\begingroup$ 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
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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, $$ \begin{align} E(Y) &= E(X_1 + \cdots + X_n) \\ &= E(X_1) + \cdots + E(X_n) \\ &= \frac{1}{n^2} + \cdots + \frac{1}{n^2} \\ &= \frac{n}{n^2}. \end{align} $$
answered Dec 13 '13 at 3:51
