Probability dice flip Central Limit Theorem problem

Question

Suppose you flip a fair dice 300 times. Let $X$ be the number of times a $6$ was thrown. What is the probability that $X$ is greater than 60?

How can I start with such a problem?

Answer 1

Let's take $X_i$ to be $1$ if you got $6$ for the $i$-th dice.

$E(X_i)=\frac{1}{6}, Var(X_i)=\frac{1}{6}-\frac{1}{36}=\frac{5}{36}$

Now we are looking for $X=\sum_{i=1}^{300}X_i$

$E(X)=E(\sum_{i=1}^{300}X_i)=\sum_{i=1}^{300}E(X_i)=50$

$Var(X)=Var(\sum_{i=1}^{300}X_i)=\sum_{i=1}^{300}Var(X_i)=300\cdot\frac{5}{36}=41\frac{2}{3}$

Now we can start playing with the central limit theorem.

When $n\rightarrow \infty $ you get $\frac{\frac{1}{n}\sum_{i=1}^{n}X_i-E(X_i)}{\sigma/\sqrt(n)}\rightarrow z$ when $z$ is a normal distribution.

So $P(X>60)=P(X-E(X)>10)=P(\frac{X}{n}-E(X_i)>\frac{10}{n})= $

$=P(\frac{\frac{x}{n}-E(X_i)}{\frac{5}{36}/\sqrt(n)}>\frac{10}{\frac{5}{36}\sqrt(n)})= P(\frac{\frac{x}{n}-E(X_i)}{\sigma/\sqrt(n)}>4.157)=1- \phi(4.157)$

So I ended up calculating $Var(X)$ for free.

I didn't justify every step as you suppose to, but if there is any step you don't understand you are welcome to ask.

And I might have some calculation errors, but the concept should be correct.