How to prove convergence in probablity Let $Y_1$, $Y_2$, ... be a sequence of random variables such that 
$P(Y_n=\frac{1}{n})=1-\frac{1}{n^2}$  and $P(Y_n=n)=\frac{1}{n^2}$.
Does $Y_n$ converge in probability?
I am stuck because I don't know how to deal with RV with two outcomes. And I am confused with the different probability in which $Y_n=1/n$ and $Y_n=n$.
I know how to prove convergence with a single $Y_n$ to Y. But really stuck in this one.
so 
$$
Y_1=\begin{cases} 
1 &\mbox{p }  = 0 \\ 
1 &\mbox{p }  = 1
\end{cases} $$
$$
Y_2=\begin{cases} 
2 &\mbox{p }  = \frac{1}{4} \\ 
\frac{1}{2} &\mbox{p }  =  \frac{3}{4}
\end{cases} $$
and then I don't know how to prove whether $Y_n$ converges or not in probability.
any help will be great. thanks
 A: If we redesign your sequence as $Y_1=1$, and everything else remains the same, the contradiction disappears. Then, if $Y_n$ converges, it should converge to $0$. Here, we'll use the definition of convergence in probability we need to show the following:
$$\lim_{n\rightarrow\infty}P(|Y_n-0|>\epsilon)=\lim_{n\rightarrow\infty} P(Y_n>\epsilon)=0$$
$Y_n$ is either $n$ or $1/n$. Now, let's fix some $n$, and calculate the above probability: 
$$\begin{align}P(Y_n>\epsilon)&=P(Y_n>\epsilon|Y_n=n)P(Y_n=n)+P(Y_n>\epsilon|Y_n=1/n)P(Y_n=1/n)\\&=P(n>\epsilon)\frac{1}{n^2}+P(1/n>\epsilon)\left(1-\frac{1}{n^2}\right)\\&=\mathbb{I}(n>\epsilon){1\over n^2}+\mathbb{I}(1/n>\epsilon)\left(1-\frac{1}{n^2}\right)\end{align}$$
Now, we take the limit: 
$$\begin{align}\lim_{n\rightarrow\infty} P(Y_n>\epsilon)&=\underbrace{\lim_{n\rightarrow\infty}\mathbb{I}(n>\epsilon)}_{1}\underbrace{\lim_{n\rightarrow\infty}\frac{1}{n^2}}_0+\underbrace{\lim_{n\rightarrow\infty}\mathbb{I}(1/n>\epsilon)}_0\underbrace{\lim_{n\rightarrow\infty}\left(1-\frac{1}{n^2}\right)}_1\\&=0\end{align}$$
which is what we needed to prove.
