# A 6-sided fair die is rolled 24 times. Let $X$ is the sum of all results. Find the probability $P(X \geq 86)$.

How can I proceed with this problem?

I started by finding z score in P:

$$P(X ≥ 86) = P(Z > (86-24)/σ)$$. Is that correct way?

• Are you looking for an exact computation or an approximation? For an approximation you can approximate with a normal distribution, but if you are looking for an exact answer, then i suspect that the calculation is quite cumbersome to do by hand, but possible if you know some basic programming skills. May 14, 2020 at 10:52
• @LeanderTilstedKristensen yeah approximation to find P(X ≥ 86) May 14, 2020 at 10:55
• 24 is $n$, you need $n \mathbf{E}X_1$ in the numerator and $\sqrt{n} \sigma$ in denominator
– Alex
May 14, 2020 at 11:17

Yes you can approximate the distribution of $$Z = \frac{X-\mathbb{E}[X]}{\sqrt{Var(X)}}$$ with a $$N(0,1)$$ distribution, but you would need to calculate both $$\mathbb{E}[X]$$ and $$Var(X)$$ in order to do this. It is useful to note, that since $$X$$ can be written as a sum of i.i.d. variables, we can calculate $$\mathbb{E}[X] = \sum_{i=1}^{24} \mathbb{E}[X_i] = 24 \mathbb{E}[X_1]$$ where $$\mathbb{E}[X_1]$$ is expected value of one dice roll, and similarly by properties of independent variables we can calculate $$Var(X) = 24Var(X_1)$$ where $$Var(X_1)$$ is the variance of a single dice roll. Once you have computed $$\mathbb{E}[X_1]$$ and $$Var(X_1)$$ you should be able to solve the problem.

Since you asked only for hints, here it is:

First step is to find out the sample space. Here are set of events:

(1,1,1... upto 24 times)

(2,2,2... upto 24 times)

_ _ _ _ _ ... 24 places

In each place you can put one out of 6 digits Therefore sample space is $$6^{24}$$

Now you have to see the digit 86.

Think of it as you have 86 envelops and 24 boxes. You can put at most 6 envelopes in one box and at least one envelope in one box.

The no. of positive integral solutions of $$x_1 + x_2$$ + ... upto $$x_r$$ = n is $$^{n-1}C_{r-1}$$