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.

However much of this work has already been done when studying Binomial Distribution, giving you a much easier tool to use.

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.


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