# probability, independent or dependent

#### 8Julio

##### New Member
Hello.

I roll a dice twice. Let's say that A = odd number on first, B = even number on second, C = sum is odd.
Now, I need to calculate P(A), P(B), P(C) and P(A cut B).

Let's start, I think these three events are independent on each other.
And thus these are ( I think ) relevant,

P(A cut B) = P(A) P(B),
P(A cut B) = P(A) P(C),
P(B cut C) = P(B) P(C),
P(A cut B cut C) = P(A) (B) P(C)

cut means intersection, as you know.
Because there are numbers 1-6,
A = 1/6, B = 1/6, C = 1/6
And therefore
P(A) = P(A) P(B) = 1/6 * 1/6 = 1/36
P(B) = P(A) P(C) = 1/6 * 1/6 = 1/36
P(C) = P(B) P(C) = 1/6 * 1/6 = 1/36
P(A cut B) = P(A cut B) = P(A cut C) = P(B cut C) = (1/6)^3 = 1/216

Maybe this is not this easy? I don't know have I done this right?

#### BGM

##### TS Contributor

A = odd number on first, B = even number on second, C = sum is odd.
I think these three events are independent on each other.
The assumption that A and B are independent is fair enough. C is pairwisely independent with A and B respectively but that is not immediate. Also note that they are not jointly/mutually independent.

Because there are numbers 1-6,
and the dice is fair so you have a discrete uniform distribution. But the events A, B, C are not asking you the probability of obtaining a specific number. So why 1/6?

#### 8Julio

##### New Member
I try again.

Because there are odd or even numbers 1-3,
A = 1/3, B = 1/3, C = 1/3

And therefore

P(A) = P(A) P(B) = 1/3 * 1/3 = 1/9 ( OR 1/3 + 1/3? )
P(B) = P(A) P(C) = 1/3 * 1/3 = 1/9
P(C) = P(B) P(C) = 1/3 * 1/3 = 1/9

P(A cut B) = P(A cut B) = P(A cut C) = P(B cut C) = (1/3)^3

#### BGM

##### TS Contributor
I think you need to start from the very basic concept before you blindly applying those properties...

In a dice, how many faces are odd? how many are even? With the equally-likely assumption (the dice is fair), what is the probability of getting odd? and even?

#### 8Julio

##### New Member
Yes... of course... I was too hasty. Probability to get odd is 1 / 2 or even is 1/ 2.

Because there are odd or even numbers
A = 1/2, B = 1/2, C = 1/2

And therefore

P(A) = P(A) P(B) = 1/2 * 1/2 = 1/4 ( OR 1/2 + 1/2? )
P(B) = P(A) P(C) = 1/2 * 1/2 = 1/4
P(C) = P(B) P(C) = 1/2 * 1/2 = 1/4

P(A cut B) = P(A cut B) = P(A cut C) = P(B cut C) = (1/2)^3

#### BGM

##### TS Contributor
I don't know why you keep saying $$P(A) = P(A)P(B)$$ which obviously does not hold unless $$P(B) = 1$$.

The definition of independence is $$P(A \cap B) = P(A)P(B)$$

And I think you can solve all with ease.

#### 8Julio

##### New Member
Okay, I try again.

P(A) = 1/2
P(B) = 1/2
P(C) = 1/2

P(A) cut P(B) = 1/2 * 1/2 = 1/4

#### 8Julio

##### New Member
I make a little adjustment here.

P(A) = 1/2
P(B) = 1/2
P(C) = 1/2

P(A cut B) = 1/2 * 1/2 = 1/4

Are these now correct?