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Let us roll a six-sided dice n times. We suppose that the tosses are independent and random. Let $X_i$ be a random variable that takes $0$ if the number obtained belongs to {1, 2} or $1$ if the result belongs to {3, 4}, or $2$ if the number belongs to {5, 6}.

The questions is which probability distribution is this for $X_i$ / $1 \leq i \leq n$ ?

I am thinking this is not Binomial, Geometric, or Poisson, because there is no success/failure. I am just confused.

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    $\begingroup$ Hello and welcome to MathStackExchange. In addition to your question, you should also add your own thoughts, work and attempts at solving it. People are more willing to help if they see that you actually tried yourself. $\endgroup$ Commented Feb 22, 2022 at 7:51
  • $\begingroup$ It has six sides $\endgroup$
    – Papa
    Commented Feb 22, 2022 at 8:00
  • $\begingroup$ I have corrected it. $\endgroup$
    – Papa
    Commented Feb 22, 2022 at 8:02
  • $\begingroup$ Look up the multinomial distribution $\endgroup$ Commented Feb 22, 2022 at 10:45
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    $\begingroup$ Whether or not it is mentioned in your syllabus, it is multinomial. But it is a special case because the probability of each value of $X_i$ is equal. Are you interested in the sum $X_1 + X_2 + \cdots + X_n$? You did not say so. $\endgroup$
    – David K
    Commented Feb 22, 2022 at 13:27

3 Answers 3

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Not exactly sure what the whole thing about the $\text{Toss}_i$-s being independent is for, but clearly each $X_i$ is discrete uniform on $\{0,1,2\}$.

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  • $\begingroup$ Toss just throw in this context. How would you determine the distribution in this case with parameters? $\endgroup$
    – Papa
    Commented Feb 22, 2022 at 8:07
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If you call the number of times $\{1,2\}$ comes up in $n$ throws as $X,$

$P(X) = (\frac13)^i*(\frac23)^{n-i}\;,\;1\leq i\leq n$ which becomes a binomial distribution

You will have parallel distributions for $Y$ and $Z$

These are binomial distributions which you have already learnt.

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You have answered the question in your post, but I will rewrite it in terms of the explicit distribution function of $X_1$ for clarity: $$ \mathbb P(X_1\leqslant t) = \frac13\cdot\mathsf 1_{[0,1)}(t) + \frac23\cdot\mathsf 1_{[1,2)}(t) + \mathsf 1_{[2,\infty)}(t),\ t\in\mathbb R. $$

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  • $\begingroup$ You haven't clarified what you mean, which distribution is this? $\endgroup$
    – Papa
    Commented Feb 23, 2022 at 12:36
  • $\begingroup$ @jii I explicitly wrote the cumulative distribution function, perhaps you are not understanding the notation? Here $\mathsf 1$ denotes the indicator function. For example: en.wikipedia.org/wiki/Indicator_function $\endgroup$
    – Math1000
    Commented Feb 25, 2022 at 12:35

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