# Dice problem. Related to probability theory

As a novice in probability theory, in this question I got stuck:

Suppose that a die is loaded so that each of the numbers $1, 2, 3, 4, 5,$ and $6$ has a different probability of appearing when the die is rolled. For $i = 1, . . . , 6$, let $\Bbb{P}(i)$ denote the probability that the number $i$ will be obtained, and suppose that: $$\Bbb{P}(i) = \begin{cases} 0.11 & i=1 \\ 0.30 & i=2 \\ 0.22 & i=3 \\ 0.05 & i=4 \\0.25 & i=5 \\ 0.07 & i=6 \end{cases}$$

Suppose also that the die is to be rolled $40$ times. Let $X_1$ denote the number of rolls for which an even number appears, and let $X_2$ denote the number of rolls for which either the number 1 or the number 3 appears.

Find the value of $$\Bbb{P}(X_1 = 20 \space\ and \space\ X_2 = 15)$$.

My working: I have the following intuition:I might have to use Identity function.

I define: $$\Bbb{I}_{X,i} = \begin{cases} 1 & \text{even number appears on the i th roll} \\ 0 & \text{odd number appears on the i th roll}\end{cases}$$ $$\Bbb{I}_{Y,i} = \begin{cases} 1 & \text{1 or 3 appears on the i th roll} \\ 0 & \text{other numbers appears on the i th roll}\end{cases}$$ $$\Bbb{E}(\Bbb{I}_{X,i})= \Bbb{P}(2)+\Bbb{P}(4)+\Bbb{P}(6) \space\text{and}\space\Bbb{E}(\Bbb{I}_{Y,i})= \Bbb{P}(1)+\Bbb{P}(3)$$

Also $$X_1=\sum_{i=1}^{40} \Bbb{I}_{X,i} \space\text{and}\space X_2=\sum_{i=1}^{40} \Bbb{I}_{Y,i}$$ I feel that this proceduer has to give a solution but I am not being able to proceed. Will someone help me proceed?

$\Bbb{EDIT}$:

Redifine $X_2$ to be the number of rolls for which either the number 2 or the number 3 appears . ( My working will be modified accordingly)

For the modified question, for each throw of the die, define the disjoint events $\{4,6\}$, $\{2\}$, $\{3\}$, and $\{1,5\}$ such that the multinomial distribution can be used to describe the outcome of the 40 throws. Let $\mathbb{P}(i,j,k,l)$ be the (multinomial) probability that we get $i$ throws that result in a four or a six, $j$ throws that result in a two, $k$ throws that result in a three, and $l$ throws that result in a one or a five. Then, the sought probability is given by summation over all $\mathbb{P}(i,j,k,l)$ such that $i+j=20$ and $j+k=15$.