# 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)

## 1 Answer

For each roll of the die, you may either get an even number, or the number one or the number three, or the number five. So you have 40 independent trials, each with three possible outcomes. The appropriate distribution for this situation is the multinomial distribution.

The multinomial distribution describes the case of independent trials which may result in several (i.e., more than two) outcomes. It is a generalization of the binomial distribution.

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 $j+k=15$.

• I have edited the question. What procedure am I now to follow? @ Apeln ....Anyways thanks for giving the catch. Dec 15, 2015 at 13:02