Convergence in probability? 

I am asking that the convergence in probability explained in the first image shows that the probability that Y-bar in the range (uY - c) to (uY + c) becomes arbitrarily close to 1 for any constant > 0 if n increases.
However, the second image from another source shows that the probability that Xn (which is equal to Y-bar?) being far from X goes to 0 for any constant > 0 as n increases. Which explanation is correct? Or is there something I misunderstand between them? Thanks a lot.
 A: The standard definition (in my experience) of $X_n\to_P X$ is that for any $\epsilon > 0,$ $P(|X_n-X|>\epsilon) \to 0,$ i.e. your second definition. (Whether it's $>\epsilon$ or $\ge \epsilon$ is immaterial.) 
However, that definition is equivalent to this one: For any $\epsilon>0,$ $P(|X_n-X| \le \epsilon) \to 1.$ The equivalence follows from the fact that $P(|X_n-X|\le \epsilon) = 1-P(|X_n-X| >\epsilon).$ This other form is is equivalent to your first definition when you substitute $c$ for $\epsilon$ and do a little unpacking.
A: The second image is the standard definition of convergence in probability which is: A sequence of random variables $X_1, X_2, \ldots, X_n$ is said to converge to the random variable $X$ in probability if for every $\epsilon > 0$, 
$$\lim\limits_{n \rightarrow \infty} P(|X_n-X|\geq \epsilon)=0 \equiv \lim\limits_{n\rightarrow \infty} P(|X_n-X|<\epsilon) = 1$$
And the first image relates to the definitions of weak law of large numbers and consistency as follows,
WLLN: Given iid random variables $Y_1, Y_2, \ldots, Y_n$ such that $E(Y_i) = \mu$ and $Var(Y_i) = \sigma^2 < \infty$. Define sequence of random variables $\overline{Y}_n = \frac{1}{n}\sum_{i=1}^{n}Y_i$, then for every $\epsilon > 0$,
$$\lim\limits_{n\rightarrow \infty}P(|\overline{Y}_n-\mu| < \epsilon) = 1$$
that is $\overline{Y}_n$  converges in probability to $\mu$.
The proof is by using the Chebyshev's inequality. For fixed $\epsilon > 0$
$$P(|\overline{Y}_n-\mu| < \epsilon) = P((\overline{Y}_n-\mu)^2 < \epsilon^2) \geq 1 - \frac{E((\overline{Y}_n-\mu)^2)}{\epsilon^2} = 1 - \frac{Var\overline{Y}_n}{\epsilon^2} = 1 - \frac{\sigma^2}{n}$$
$$\implies \lim\limits_{n\rightarrow\infty} P(|\overline{Y}_n-\mu| < \epsilon) = 1$$
The property summarized by the WLLN, that a sequence of the "same" sample quantity approaches a constant as $n \rightarrow \infty$, is known as consistency.
