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?

  • 1
    $\begingroup$ 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. $\endgroup$ May 14, 2020 at 10:52
  • $\begingroup$ @LeanderTilstedKristensen yeah approximation to find P(X ≥ 86) $\endgroup$
    – HuanCarlos
    May 14, 2020 at 10:55
  • $\begingroup$ 24 is $n$, you need $n \mathbf{E}X_1$ in the numerator and $\sqrt{n} \sigma$ in denominator $\endgroup$
    – Alex
    May 14, 2020 at 11:17

2 Answers 2


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.

There is basically a ready made formulae available for this.

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

In this case, n= 86 and r=24

By this way you find the total no. of cases when the sum of outcomes equal to 86


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