# If six cubic dice are thrown (not loaded), this is a $X$ ~ $Bin(6;p)$?

## Exercise says this:

Six cubic dice are thrown (not loaded). Calling a success to get a 5 or a 6,

Calculates the probability of obtaining:

• A) Exactly three successes.
• B) A maximum of three successes.
• C) At least three successes.

## My question is this:

It is like repeating the experiment of throwing a dice, but 6 times? That is, a binomial distribution.

Or are we talking about $6 ^ 6$ possibilities and the experiment is repeated only once?

If it were the first case, my $p = 1/3$ and my $n = 6$. Being $X: "$Sometimes I get either the number 5 or the number 6$"$ I have $X$ ~ $Bin (6;1 / 3)$

Well, you may model it as a sample space with $6^6$ equally probable outcomes and try to work out the probabilities from first principles.   You can obtain the answers that way.
We have a sequence of $6$ Bernoulli trials with success rate of $1/3$ and wish to count the successes among the results.   That is indeed a Binomially distributed random variable.
$$X\sim\mathcal{Bin}(6, 1/3)\qquad\checkmark$$
Both interpretations are correct, but looking at a binomial with $n=6$ and $p-1/3$ will make your work easier.