# Simple joint probability question of two dice throw

I have a simple joint probability question I cannot understand. If I throw two dice and define the following events:

$$X$$ - number of times 4 was obtained.

$$Y$$ - number of even results obtained.

Now I need to find the joint probability function, so my problem is with $$P(x=0, y=1) = P(\text{obtaining zero times four and one time an even result})$$, so it should be equal to probability of obtaining uneven results, and probability of obtaining an even result times $$2/3$$ (since 4 was not received). From this reasoning, I think the result should be: $$\frac{3}{6} \cdot (\frac{3}{6} \cdot \frac{2}{6})$$, i.em $$1/2$$ chance to obtain uneven number times the probability to obtain an even number that is not ($$3/6 \cdot 2/6$$).

However, the correct result should be: $$\frac{2 \cdot 2 \cdot3}{6 \cdot 6}$$, and I don't understand why. I would appreciate an explanation.

EDIT: similarly, why in $$P(x=1, y=2) = P(\text{two even results, of which only one is four})$$ = $$\frac{2 \cdot (1 \cdot 2)}{6 \cdot 6}$$ and not: even ($$3/6$$) * even that is not four ($$2/6$$)?

Give the dice a number.

Let $$D_1$$ denote the result of the first and let $$D_2$$ denote the result of the second die.

Then:

$$P(X=0,Y=1)=P(D_1\in\{2,6\}, D_2\in\{1,3,5\})+P(D_1\in\{1,3,5\}, D_2\in\{2,6\})=$$$$2P(D_1\in\{2,6\})P(D_2\in\{1,3,5\})=2\frac26\frac36$$

Edit (after edit of question):

$$P(X=1,Y=2)=P(D_1=4, D_2\in\{2,6\})+P(D_1\in\{2,6\}, D_2=4)=$$$$2P(D_1=4)P(D_2\in\{2,6\})=2\frac16\frac26$$

• thank you very much, your explanation really helps me understand what i'm calculating and how
– q123
Commented Jan 28, 2019 at 12:43
• You are very welcome. In questions like this (you roll $n>1$ dice) it is often helpful to give the dice numbers. Commented Jan 28, 2019 at 12:45