Question regarding Probability of Dice

Question

A pair of dice is thrown 180 times in a row. Find the probability that the event is 25 or more times I should define random variable $X$ but is this binomial distribution or negative binomial distribution. I really confused the define random variable and apply the distribution function. Any idea will be appreciated.

Answer 1

Using CLT you get

$$\mathbb{P}[X\geq 25]=\mathbb{P}\left[Z\geq \frac{24.5-180/6}{\sqrt{180\cdot\frac{5}{6}\cdot\frac{1}{6}}}\right]=\mathbb{P}[Z\geq -1.1]=1-\Phi(-1.1)\approx 86.43\%$$

this because your rv is a bernulli taking values zero and one with probability $\left\{\frac{5}{6};\frac{1}{6}\right\}$ respectively (one if the result is 7 and zero if result is not 7)

The exact result with the binomial is $\approx 86.56\%$

Nice approx with CLT, a poket calculator and the paper gaussian table...

Answer 2

As lulu said in the comments, we just need to calculate the probability that a single throw of pair yields 7.

Let A denotes the incident that in one trial the total on the two dice is 7 and $B_{ij}$ denotes the incident that the number on the $j^{th}$ dice is i. Then we have $$A = \sum_{i=1}^6 AB_{i1} $$ Thus we can calculate the $\mathbb{P}$(A) \begin{equation}\begin{aligned} \mathbb{P}(A) &= \sum_{i=1}^6\mathbb{P}(AB_{i1})\\ &= \sum_{i=1}^6\mathbb{P}(A|B_{i1})\mathbb{P}(B_{i1})\\ &= \sum_{i=1}^6\mathbb{P}(B_{7-i,2}|B_{i1})\mathbb{P}(B_{i1})\\ &= \sum_{i=1}^6\mathbb{P}(B_{7-i,2})\mathbb{P}(B_{i1})\\ &= \frac{1}{6} \end{aligned}\end{equation}

So the answer of your question is $\sum_{i=25}^{180}\binom{180}{i}(\frac{1}{6})^{i}(\frac{5}{6})^{180-i}$, which is the same as yours.