How to prove $Z_{n}$ converges to $p^{2}$ in probability?

Question

I'm solving this exercise

and get stuck at Question 3b, i.e, Show that $Z_{n}$ converges to $p^{2}$ in probability.

Could you please shed me some light on 3b?

My attempt:

b. For $\varepsilon > 0$, we have $\mathbb P(|Z_n -p^2| > \varepsilon) = \mathbb P (Z_n \in (p^2 - \varepsilon, p^2 + \varepsilon))$.

c. We have $\mathbb E(Y_1) = \mathbb E(X_1 X_2) = \mathbb E(X_1) \mathbb E(X_2) = p^2$. Thus $$\begin{aligned} \mathbb E(Z_n - p^2)^2 &= {(\mathbb E(Z_n)-p^2 )}^2 +\operatorname{Var}(Z_n) \\ &= {(\mathbb E(Y_1)-p^2 )}^2 + \operatorname{Var}(Z_n)\\ &= {(p^2-p^2 )}^2 + \frac{p^{2}\left(1-p^{2}\right)}{n}+\frac{2(n-1)\left(p^{3}-p^{4}\right)}{n^{2}} \\ &=\frac{p^{2}\left(1-p^{2}\right)}{n}+\frac{2(n-1)\left(p^{3}-p^{4}\right)}{n^{2}} \end{aligned}$$

Hence $$\lim_{n \to \infty} \mathbb E(Z_n - p^2)^2 = 0$$

Answer 1

From @Calvin Lin's comment

Use part a which shows that the variance tends to $0$. Hence $Z_n$ cannot vary too far from its mean, E.g. Chebyshev.

I post it as an answer to peacefully close this question.