# Conditional Multivariate Probability

#### oxycontiin

##### New Member
So, I got this from my textbook and it makes sense:

$$P_{Y|x}(y)= P(Y=y|X=x) = P_{X,Y}(x, y)/P_X(x)$$

But I'm trying to apply it to this question and I'm struggling. Here's the question:

Suppose a die is rolled six times. Let X be the total number of 4’s that occur and let Y be the number of 4’s in the first two tosses. Find $$P_{Y|x}(y)$$.

What I've got so far:

$$P_X(x) = ((1/6)^x)((5/6)^{6-x})(_6C_x)$$

but now I need to find $$P_{X,Y}(x,y)$$ and I'm kind of lost. What do I do next?

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#### oxycontiin

##### New Member
Thanks, I thought it looked kind of illegible. I'll rewrite it right now.

#### Dason

How are you getting $$P_X(x) = 3P_Y(y)$$?

#### oxycontiin

##### New Member
I didn't really put much thought into that one actually, it was kind of just an idea. I figured if $$P_X(x)$$ was for six rolls, then a third of that would work for two. That definitely isn't true, I'll take it out. I would have to adjust the formula for two rolls, which means changing the second exponent from 6-x to 2-y and $$_6C_x$$ to $$_2C_y$$.

#### BGM

##### TS Contributor
Let $$Z$$ be the number of 4's in the last 4 tosses. Then $$Z$$ also follows a Binomial distribution, independent of $$Y$$ and $$X = Y + Z$$.

Hence you have the result:
$$P_{X,Y}(x,y) = \Pr\{X = x, Y = y\} = \Pr\{Y + Z = x, Y = y\} = \Pr\{Z = x - y\}\Pr\{Y = y\}$$

#### oxycontiin

##### New Member
Ah, that's an interesting approach. I will take a look at that. Thanks!

Oh, it's independent. Really? Even if X and Y are dependent? I guess the last four heads have to be independent of the first two. If I'm understanding that correctly then that is very clever. I will solve right now and come back with an answer.

Okay, I took another look, but now I'm a little confused. Hopefully I can get your input again. So I solved the probability you wrote: $$Pr(Z = x-y)Pr(Y=y)$$

This is what I got:

$$P_Y(y) = (1/6)^y(5/6)^{2-y}(_2C_y)$$
$$P_Z(z) = (1/6)^z(5/6)^{4-z}(_4C_z)$$
$$Pr(Z = x-y)Pr(Y=y) = [(1/6)^z(5/6)^{4-z}(_4C_z)][(1/6)^y(5/6)^{2-y}(_2C_y)]$$
I tried simplifying, but I'm not very good at it so it could be wrong: $$(1/6)^{zy}(5/6)^{8-2z-4y+zy}(_2C_y)(_4C_z)$$

But after solving it, it seems to me that p(x,y) is different from what you wrote, since p(x,y) would be the probabilities of getting heads on any of the six rolls & getting heads on the first two rolls vs. pr(z)pr(y) which would be getting heads on the last four & getting heads on the first two. Wouldn't your probability just equate to $$P_X(x)$$?

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#### BGM

##### TS Contributor
I mean $$Y, Z$$ are independent. Of course $$X$$ is dependent on both $$Y$$ and $$Z$$.

Not sure what is giving you troubles but note that

$$\Pr\{Z = x - y\} = P_Z(x - y)$$

according to your notation and therefore there will be no $$z$$ in the later part of your expression. It is just an expression in terms of $$x$$ and $$y$$

#### oxycontiin

##### New Member
Right, I understand that Y and Z are independent. That makes sense. I think I'm just confusing myself. I'm looking at your previous post again. $$Pr(Y+Z=x, Y=y)$$ makes sense to me. But when you subtract Y I feel like it's not answering the question anymore. I agree with the equation you wrote X = Y + Z and it makes it independent, which is nice, but it was supposed to be $$P_{X,Y}(x,y)$$ and it doesn't look like $$P_{X,Y}(x,y)=P_Z(x-y)P_Y(y)$$. If it is true could you say that $$P_Y(y)=P_Y(y)$$ so $$P_X(x)=P_Z(x-y)$$? Would my answer in the previous post be correct if I accept your equation to be true? I should probably ask what Pr is. I missed the class that taught conditional probabilities for multivariate problems, which is why I'm a little behind now.

Okay I've thought about this for a while, I think I understand. You know that X = Z + Y is true and since all $$P_{X,Y}(x,y)$$ is asking for is a probability involving a relationship between X and Y then you can solve X through a combination of Y and Z and so the statement $$P_Z(x-y)P_Y(y)$$ still stands. In that case, would my answer be correct?

Thank you very much for your help.

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