Probability of rolling a $6$ between $90$ and $110$ times when rolling a die $600$ times

Question

Show that if you roll a fair die $600$ times, then with probability at least $1/6$ you will get the roll $6$ between $90$ and $110$ times.

This is what I have so far

Let $E(x) = 100 \Rightarrow U(x) = 100 \cdot 1/6 = 600$

But, I don't know what more I need to show?

Thanks for your help.

Answer 1

You have a binomial distribution so you need to demonstrate

$$\sum_{k=90}^{110}\binom{600}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{n-k}\ge \frac{1}{6}$$

You can either demonstrate this directly or by arguing from the expected value and the variance using a normal distribution approximation.

Since the probability is in fact $\gt\frac{3}{4}$ you do not need to add up all of these terms. $\sum_{k=98}^{101}$ is $\gt\frac{1}{6}$

Answer 2

It's a binomial distribution. The mean is $100$, and the standard deviation is calculated as follows:

$$\sigma^2=np(1-p)$$ $$\sigma^2=600*\frac{1}{6}\frac{5}{6}=83.3333$$ $$\sigma\approx9.12$$

So the range for this interval is found as:

$$[\mu-\sigma,\mu+\sigma]$$