im confused about what this problem is asking: A die is cast 5 times, event X1 is a success for {1,2,3} and X2 is success for {4,5}. find: $$P(X_1=2,X_2=1)$$ is this asking for the probabilities of X1 and X2?
this is what i have: $$P(X_1)= {{5}\choose{2}}(1/2)^2(1/2)^3=5/16 $$ $$P(X_2)= {{5}\choose{1}}(1/3)^1(2/3)^4=80/243 $$ $$P(X_1=2,X_2=1)=\frac{5}{16}\frac{80}{24}=\frac{25}{243}$$
In each trial you have three mutually exclusive and exhaustive events. $$E_1=\{1,2,3\}, E_2=\{4,5\}, E_3=\{6\}$$
Let $p_1,p_2,p_3$ be the respective probabilities for these events on any particular trial.
Let $X_1,X_2,X_3$ be the respective counts of occurrences of these events in five trials.
These random variables will follow a multinomial distribution; which is an extension of a binomial distribution.
$$\mathsf P(X_1{=}x,X_2{=}y, X_3{=}(5-x-y)) ~=~ \dfrac{5!}{x!\,y!\,(5-x-y)!}\; p_1^{\raise{0.5ex}x}\,p_2^y\,p_3^{5-x-y}$$
We turn it back over to you.