Let $X$ be the number of $1$s and let $Y$ be the number of $2$s obtained. What is the joint PMF of $X$ and $Y$.

Question

I was reading Introduction to Probability, 2nd Edition and the following problem appears on page 128:

Consider four independent rolls of a 6-sided die.

Let $X$ be the number of $1$s and let $Y$ be the number of $2$s obtained.

What is the joint PMF of $X$ and $Y$?

My reasoning to solve the problem is as follows:

We need $P_{X,Y}(x, y)$.

Let us compute the probability of having $x + y \ $ $1s$ or $2s$ first, it is: $${}_4 \mathrm{ C }_{x+y} \cdot \left(\frac{2}{6}\right)^{x+y} \cdot \left(\frac{4}{6}\right)^{4-(x + y)}$$

Now out of the $x + y$ $1$s or $2$s the probability of having $x \space 1s$ is:

$${}_{x + y} \mathrm{ C }_{x} \cdot \left( \frac{1}{2} \right)^{x + y}$$

And the final answer is the product of the two probabilities.

Is my reasoning wrong? and if so then what is my mistake?

Answer 1

The two marginals are binomial but they are not independent thus to show the joint distribution is better to use a table-representation.

As an example, here below is the joint distribution related to 2 independent rolls of a die... you can easy do the same work for $n=4$. The resulting table is a triangular matrix...

	Y=0	Y=1	Y=2	Total
X=0	p(0;0)	p(0;1)	p(0;2)	25/36
X=1	p(1;0)	p(1;1)	0	10/36
X=2	p(2;0)	0	0	1/36
Total	25/36	10/36	1/36	1/36

I think it is easy for you to calculate the probabilties in the body of the table which have to sum both per column and per row to the binomial probabilities