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We roll a non-standard dice three times (chance of getting a tail is $25\%$, for head it's $75\%$). Let $X$ be a number of tails that we got from first two throws, and $Y$ - number of tails gotten from last two throws. Find joint distribution of the random vector $(X,Y)$. Are variables $X$ and $Y$ independent?

I guess that first step would be finding $\mathbb{P}\left(X=\dots\right)$ and $\mathbb{P}\left(Y=\dots\right)$.

So:

$\mathbb{P}\left(X=0\right)={2\choose 0}\cdot\left(\frac{1}{4}\right)^0\cdot\left(\frac{3}{4}\right)^2=\frac{9}{16} \\ \mathbb{P}\left(X=1\right)={2\choose 1}\cdot\left(\frac{1}{4}\right)^1\cdot\left(\frac{3}{4}\right)^1=\frac{6}{16}\\ \mathbb{P}\left(X=2\right)={2\choose 2}\cdot\left(\frac{1}{4}\right)^2\cdot\left(\frac{3}{4}\right)^0 =\frac{1}{16}\\ $

For $\mathbb{P}\left(Y=\dots\right)$ it is exactly the same.

But.. What should I do next?

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    Is $Pr(X=2~\&~Y=2)=Pr(X=2)\cdot Pr(Y=2)$? What is $Pr(X=a~\&~Y=b)$ for each choice of $(a,b)$? – 2017-01-31
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    @JMoravitz This would be correct if $X$ and $Y$ were independent, right? Hmmm, as for the second question - maybe something like this: $Pr(X=a|Y=b)=\frac{Pr(X=a,Y=b)}{Pr(Y=b)}$ And from this formula I get what you ask about. – 2017-01-31
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    I used an ampersand symbol by which I meant an "*and*", not a "*given*"., reworded, $Pr((X,Y)=(a,b))$ or $Pr(X=a\wedge Y=b)$, or as you seem to prefer $Pr(X=a,Y=b)$. The problem asks you to write out the joint distribution explicitly. – 2017-01-31
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    @JMoravitz I already figured that out, since it is my task. I am having a problem how to accomplish that. But let's say that I want to find the probability of $X=0$ and $Y=0$, $Pr(X=0, Y=0)$. The only possible situation is when I get 3 heads in 3 flips. So is $Pr(X=0, Y=0)$ equal to ${3\choose 0}\cdot\left(\frac{1}{4}\right)^0\cdot\left(\frac{3}{4}\right)^3=\frac{27}{64}$ ? – 2017-01-31
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    Correct. Note that $X=1$ and $Y=1$ corresponds either to the scenario of HTH or to the scenario of THT. That is perhaps the trickiest one to calculate. Others are easier because knowledge of the one forces the other, like X=2 and Y=1 must occur as TTH, etc... – 2017-01-31
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    @JMoravitz ok, that's a progress. Tell me, please, if this is correct: $Pr(X=1,Y=1)=\frac{3}{4}\cdot\frac{1}{4}\cdot\frac{3}{4}+\frac{1}{4}\cdot\frac{3}{4}\cdot\frac{1}{4}=\dots$ and if it is ok, I think I am good to go. Oh, and one more question - to check if $X,Y$ are indepedent I need to calculate, for example, if $Pr(X=0,Y=0)=Pr(X=0)Pr(Y=0)$? – 2017-01-31
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    @SantaXL That is correct. But more, you need to determine whether it holds for all values. (Notably, look at $P(X=0, Y=2)$) – 2017-01-31

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