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?