Convergence in Probability Consider a sequence of $N$ Bernoulli trials with, with probability of success denoted by $p$, and let $X$ be the number of successes. Show that as $N\rightarrow\infty$, $\frac{X}{N}$ converges in probability to $p$.
I am not sure how to tackle this but I think that as if this occurs then by definition $P\left(\frac{X}{N} - p \geq \varepsilon\right)= 0$, as $N\rightarrow\infty$. How would I show this is through?
 A: $\newcommand{\var}{\operatorname{var}}\newcommand{\E}{\operatorname{E}}$Markov's inequality says if $Y$ is a non-negative random variable and $y>0$, then
$$
\Pr(Y\ge y) \le \frac{\E(Y)} y.
$$
(See the bottom of this posting for a proof of Markov's inequality.)
Applying Markov's inequality to $(X-\mu)^2$, where $\mu=\E X$, we get $\Pr((X-\mu)^2\ge d^2)\le \dfrac{\sigma^2}{d^2}$ where $\sigma^2=\var X$. From that we deduce Chebyshev's inequality:
$$
\Pr\left(|X-\mu|\ge d\right)\le \frac{\sigma^2}{d^2}.
$$
For the random variable $X$ defined in the question we have
$$
\E \frac X N = p\qquad\text{and}\qquad \var \frac X N = \frac{p(1-p)} N.
$$
So by Chebyshev's inequality,
$$
\Pr\left( \left| \frac X N - p \right| \ge \varepsilon \right) \le \frac{p(1-p)}{\varepsilon^2 N} \to 0 \text{ as }N\to\infty.
$$
Proof of Markov's inequality:
Let $W=\begin{cases} y & \text{if }Y\ge y, \\ 0 & \text{otherwise}. \end{cases}$
Then $0\le W\le Y$, so $\E W\le \E Y$.  And $\E W = y\Pr(Y\ge y)$. ${}\qquad\blacksquare$
A: Recall that the Weak Law of Large Numbers (see Wikipedia) asserts that if $Y_1,Y_2,Y_3,\ldots$ are independent realizations of some random variable $Y$ such that $\mu=\mathbb E[Y]$, then the sample average
$$\sum_{i=1}^N\frac{Y_i}N$$
converges to the theoretical mean $\mu$ in probability.
In the context of your specific question,
find a random variable $Z$ whose mean is your probability $p$ of success,
and such that the sample average of trials of i.i.d. copies of $Z$ will be equal to $X/N$.
