Conditional probability problem

[biometrix]

Hello friends.

I'm trying to work out this simple example and I'm stuck on the priciples. The example is:

Suppose X is a random variable for a rolling 6-side dice. Suppose A the event that an outcome is even. What is the conditional probability P(X|A)?

Without much calculation and just some thinking the answer comes to be 1/3.

My question arises when we write down the formula:

[MATH] P(X|A) = \frac{P(X=k, A)}{P(A)} \Rightarrow P(X=K, A) = P(X|A)\cdot P(A) \Rightarrow P(X=k, A) = \frac{1}{3}\cdot \frac{1}{2} = \frac{1}{6}[/MATH].

What does it mean that the probability of the event that both X and A happen together, is 1/6? Is it correct?

This leads to things like:

[MATH]P(X \cup A) = P(X) + P(A) - P(X \cap A) = \frac{1}{6} + \frac{1}{2} - \frac{1}{6} = \frac{1}{2}.[/MATH]
But the union of A and X is X, whose probability is 1/6.

I got it wrong somewhere, thanks if you can help.

 

[pka]

Hello friends.

I'm trying to work out this simple example and I'm stuck on the priciples. The example is:

Suppose X is a random variable for a rolling 6-side dice. Suppose A the event that an outcome is even. What is the conditional probability P(X|A)?
Without much calculation and just some thinking the answer comes to be 1/3.
My question arises when we write down the formula:
[MATH] P(X|A) = \frac{P(X=k, A)}{P(A)} \Rightarrow P(X=K, A) = P(X|A)\cdot P(A) \Rightarrow P(X=k, A) = \frac{1}{3}\cdot \frac{1}{2} = \frac{1}{6}[/MATH].

Say we want \(\mathcal{P}(X=6|A)\). What is \(\dfrac{\mathcal{P}(X=6\cap A)}{\mathcal{P}( A)}\)​

Think about the conditioning. We are given (told) the die is even. So what is the probability it is a six?​

EVEN? Thus \(X\in\{2,4,6\}\) what chance does it have of being a six?​

​

BTW: \(\dfrac{\frac{1}{6}}{\frac{1}{2}}=\dfrac{1}{3}\)​

 

[Steven G]

1st, dice is the plural of die. That is you have two dice but she has one die.

Do you understand what X means? Personally I thing you don't. It appears to me that you think X is some number, like the face on a die.

X is defined to be the outcome of the toss of a die. So X can take on the values 1,2,3,4,5 and 6.

So talking about p(X|A) is meaningless!

You can say that p( X=1 | A) = 0.
p(X=2 | A) = 1/3
p( X =3 | A) = 0.
...

 

[Steven G]

Say we want \(\mathcal{P}(X=6|A)\). What is \(\dfrac{\mathcal{P}(X=6\cap A)}{\mathcal{P}( A)}\)​

Think about the conditioning. We are given (told) the die is even. So what is the probability it is a six?​

EVEN? Thus \(X\in\{2,4,6\}\) what chance does it have of being a six?​

​

BTW: \(\dfrac{\frac{1}{6}}{\frac{1}{2}}=\dfrac{1}{3}\)​

Why is \(X\in\{2,4,6\}\), why not \(X\in\{1,2,3,4,5,6\}\)?

 

[pka]

Why is \(X\in\{2,4,6\}\), why not \(X\in\{1,2,3,4,5,6\}\)?

The event \(\mathcal{A}\) is that the die shows an even number.(as posted "A is the event that an outcome is even").
So \(\mathcal{P}(\mathcal{X}|\mathcal{A})\) means \(\mathcal{X}\in\{2,4,6\}\)

 

[Steven G]

The event \(\mathcal{A}\) is that the die shows an even number.(as posted "A is the event that an outcome is even").
So \(\mathcal{P}(\mathcal{X}|\mathcal{A})\) means \(\mathcal{X}\in\{2,4,6\}\)

What is wrong with p(X=3 | A) = 0? That is why can't X=3? After all X is defined as a random variable for rolling a 6-side die

 

[pka]

What is wrong with p(X=3 | A) = 0? That is why can't X=3? After all X is defined as a random variable for rolling a 6-side die

The is absolutely nothing wrong with that. \(\mathcal{P}(\mathcal{X}=3|\mathcal{A})=0\).
It is after all conditional probability. Given the condition that the die has come up even then we know that that \(\mathcal{X}\ne 3\)
The Rev. Mr Bayes who first proved this idea, would be absolutely aghast at what we have done with it.
All he said is that if we know some event happens in a population then the probability of it happening in general is the ratio of it happening in that population divided by the probability of it being in that population.

 

[biometrix]

1st, dice is the plural of die. That is you have two dice but she has one die.

Do you understand what X means? Personally I thing you don't. It appears to me that you think X is some number, like the face on a die.

X is defined to be the outcome of the toss of a die. So X can take on the values 1,2,3,4,5 and 6.

So talking about p(X|A) is meaningless!

You can say that p( X=1 | A) = 0.
p(X=2 | A) = 1/3
p( X =3 | A) = 0.
...

Thanks for your correction but I thought this was a maths and not english forum. As I've read many posts here it would take you forever to correct every post. Now, about my question I understand what X is and sorry for using bad notation thus creating confusion. Now given that you could still answer the question if you wanted, but anyway thanks for your time.