Probability of unbiased die

Question

One of the numbers 1,2,...,6 is to be chosen by casting an unbiased die.Let this random experiment be repeated five independent times.Let this random variable $X_1$ be the number of termination in the set {$x:x=1,2,3$} and let the random variable $X_2$ be the number of terminations in the set {$x:x=4,5$}.Compute $P(X_1=2, X_2=1)$. I do not know how to begin with.. please help.. I only have the pmf of x which Binomial with n=5 and p=1/6.

Answer 1

You have a multinomial distribution, with parameters $n=5, p_1=3/6, p_2=2/6$. (Pedantically, it's as trinomial distribution.)

Going back to first principles, just as a binomial distribution's probability mass function is determined by taking the probability of a obtaining so many successes and failures in a certain sequence and multiplying by the count of permutations, we apply the same method to determine the pmf of a multinomial distribution with each of $n$ trials partitioned by several types of disjoint successes and a failure.

$\begin{align} \text{Binomial Distribution} & \quad\mathcal{Bin}(n; p_1) \\[1ex] \mathsf P(X_1\!=\!x_1) & = \frac{n!}{x_1!\;(1-x_1)!}\;p_1^{x_1}\;(1-p_1)^{n-x_1} \\[2ex] \text{Multinomial Distribution} & \quad\mathcal{Mult}(n;p_1,p_2,1-p_1-p_2) \\[1ex] \mathsf P(X_1\!=\!x_1, X_2\!=\!x_2) & = \frac{n!}{x_1!\;x_2!\;(1-x_1-x_2)!}\;p_1^{x_1}\;p_2^{x_2}\;(1-p_1-p_2)^{n-x_1-x_2} \end{align}$