Computer discrete marginal pdf

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I don't understand the solution to the second part of this problem. The problem states:

Consider the experiment of tossing a fair coin three times. Let $X$ denote the number of heads on the last flip, and let $Y$ denote the total number of heads on the three flips. Find $p_{X,Y} (x, y)$.

This is easy enough. The answer is:

(X, Y)  PXY
-----------
(0,0)   1/8
(0,1)   2/8
(0,2)   1/8
(0,3)   0
(1,0)   0
(1,1)   1/8
(1,2)   2/8
(1,3)   1/8

What I don't understand is the second part. It says:

Find the marginal pdfs of X and Y for the joint pdf

The answer is: $P_X (x) = 1/8 + 2/8 + 1/8 = 1/2, x = 0,1$ $P_Y(0) = 1/8, P_Y(1) = 3/8, P_Y(2) = 3/8, P+Y(3) = 1/8$

I don't understand two things: Why is the format for the answer of $P_X$ and $P_Y$ different? It looks to me like they're just doing different things for Px and Py to reach the answer. For Py, they look at each possible value (y = 0,1,2,3) and they state the probability of each. For Px, they add some probabilities. I don't know why it equals 4/8 and not 8/8 - using the method they did for Py, I should reach 8/8. Can someone explain to me?

Thanks

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The notation for $P_X$ is just simplifying the process - $P_X(0) = \frac{1}{8} + \frac{2}{8} + \frac{1}{8} + 0 = \frac{1}{2}$, and $P_X(1) = 0 + \frac{1}{8} + \frac{2}{8} + \frac{1}{8} = \frac{1}{2}$. They're the same sum, so instead they've been collapsed into $P_X(x) = \frac{1}{2},\ x = 0, 1$, meaning that you can substitute either value of $x$ into the formula and it's true.

For the $P_Y$ case, they just skipped showing what things got added together. So, for example, $P_Y(1) = \frac{2}{8} + \frac{1}{8} = \frac{3}{8}$.