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Roll a $4$-sided die. Whatever value appears, flip exactly that many fair coins. [For instance, if the die shows $3$, then flip $3 $fair coins.] Let $X$ denote the number of heads that appear on the coins; let $Y$ denote the number of tails that appear on the coins. Find the mass $pX,Y (x,y)$ for all integers $x \ge 0$ and $y \ge 0$ that satisfy $1 \le x+y \le 4$. [Hint: There are only fourteen such values of the pair $(x, y)$.]

I need help finding $p(1,1)$ $p(2,1)$ $p(3,1)$ and $p(2,2)$. For example, I thought $p(1,1)$ implies you roll a $2$ so the probability is $\left(\frac{1}{4}\right) \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) = \frac{1}{16}$ but the correct answer is $1/8$.

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    $\begingroup$ To compute $p(1,1)$: first, you must roll a $2$, a $\frac 14$ event. Then you need to split your two tosses, either getting $HT$ or $TH$, a $\frac 12$ event. Thus $p(1,1)=\frac 14\times \frac 12=\frac 18$. $\endgroup$ – lulu May 7 '17 at 0:39
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    $\begingroup$ Note: your calculation correctly computes the chances of rolling a $2$, then getting $HT$ in that order, but that's not what you were asked. All you want is the total number of Heads and Tails, so you have to consider both $HT$ and $TH$. $\endgroup$ – lulu May 7 '17 at 0:41
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    $\begingroup$ @lulu this should be an answer. $\endgroup$ – Oscar Lanzi May 7 '17 at 2:27
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Your error, as noted in the comments, was neglecting that one head and one tail could be formed in two ways, vis: $\rm HT, TH$.   So $p_{X,Y}(1,1)= \tfrac 14 \tfrac {2~~}{2^2}=\tfrac 18$.

Indeed, the sum of heads and tails, $X+Y$, is uniformly distributed over $\{1,2,3,4\}$, and the count of heads is binomially distributed when conditioned on that sum.

$\begin{align}p_{X,Y}(x,y) ~&=~ p_{X+Y}(x+y)\cdot p_{X\mid X+Y}(x\mid x+y)\\[1ex] &=~ \tfrac 14\cdot \tbinom{x+y}x2^{-(x+y)} ~\big[(x,y)\in S\big]\\[1ex] &=~ \tbinom{x+y}x 2^{-(2+x+y)} ~~~\big[(x,y)\in S\big]\end{align}$

So again, $p_{X,Y}(1,1)= \tbinom 2 1 2^{-4} = 8^{-1}$

Now, use this to evaluate the probability mass at all fourteen of the values in the support set, $S$.


Hint: Symmetry will save some effort if doing so by hand.

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