Consider rolling a 6-sided dice {1,2,3,4,5,6}. Find the probability that out of n trials, we throw r sixes.
I need to use a binomial event. Any hints?
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1$\begingroup$ Hint: formula Binomal distribution $\endgroup$– imranfatFeb 3, 2014 at 16:24
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$\begingroup$ Success: a $6$; Failure, anything else. We want the probability of $r$ successes (and therefore $n-r$ failures) in $n$ trials. $\endgroup$– André NicolasFeb 3, 2014 at 16:41
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$\begingroup$ Remember a binomial random variable is just a counting variable. It counts how many times an event happens over a finite amount of trials. So your counts will be how many sixes you roll with a probability of success being $\frac{1}{6}$, thus looking at $Bin(n,\frac{1}{6})$ $\endgroup$– KamsterJul 4, 2014 at 23:16
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1 Answer
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Further to the hints given in the comments: you wish to have a way of modelling the probability that after rolling the dice $n$ times there will be $r$ sixes. We would use the Binomial distribution to model this so long as we ensure the the trials are independent (that is the probability of rolling a $6$ is unaffected by the outcome of previous rolls, which of course it isn't) and that there are only two outcomes of the trial that are possible (in this case: successfully rolling a 6 or failing to roll a six).
The probability mass function for the binomial distribution is given by the formula: $$ P(X=r)={n\choose r}p^r(1-p)^{n-r} $$ where the probability of a success is $p$ (that is rolling a 6, and the probability of not rolling a 6; since there are the only two possibilities hence a binomial event) and the probability of a failure is therefore $(1-p)$. Plugging into the above formula the values from your problem that is $p=\frac{1}{6}$ givens the probability required: $$ P(X=r)={n\choose r}\left(\frac{1}{6}\right)^r\left(\frac{5}{6}\right)^{n-r}$$ If you want to form an intuition as to why the able formula works, it may help to draw a tree diagram of the possibilities to see why the $n$ choose $r$ is in it. It also may help to consider the case for small $n$ for example $n=3$, a possible outcome is SSF (in the example you've given that would be rolling a six twice and then not rolling a six) but it's apparent that this is also equivalent to SFS since it still has the same number of successes or failures, they've just happened in a different order. It turns out that there are ${3\choose2}=3$ possible different orders (the other being FSS). So in general with $n$ objects with $r$ of them being successes and $(n-r)$ being failures the number of ways for ordering the possible outcomes is $n\choose r$.
