<s>[INST] Flipping two fair coins probability function $4$ Consider flipping two fair coins. Let $X = 1$ if the first coin is heads, and $X = 0$
if the first coin is tails. Let $Y = 1$ if the two coins show the same thing (i.e., both heads
or both tails), with $Y = 0$ otherwise. Let $Z = X + Y$, and $W = XY$.
(a) What is the probability function of $Z$?

Attempt:
$P(X = 1) = 1/2$
$P(X = 0) = 1/2$
$P(Y = 1) = 1/2$ since {HH, TT, TH, HT}, 2/4 times.
$P(Y = 0) = 1/2$ (not sure bout this)
(a)
Given $X = 1$, $Z = 2,1$
Given $X = 0$, $Z = 1, 0$
$P(Z = 2) = P(X = 1) + P(Y = 1) = 1/2 + 1/2 = 1$ [Book's answer is 1/4]
$P(Z = 1) = P(X = 1) + P(Y = 0) + P(X = 0) + P(Y = 1) = 2$ [1/2 book]
$P(Z = 0) = P(X = 0) + P(Y = 0) = 1$ [1/4 book]
Could someone explain to me why they multiply instead of adding them? Thanks
 [/INST] You're on the right track. You can case out the whole function since there are only four cases.
\begin{array}{|c|c|c|c|c|c|}
\hline
Coin 1& Coin 2 & X & Y & Z & W \\ \hline
H & H & 1 & 1 & 2 & 1\\ \hline
H & T & 1 & 0 & 1 & 0\\ \hline
T & H & 0 & 0 & 0 & 0\\ \hline
T & T & 0 & 1 & 1 & 0\\ \hline
\end{array}
Note that $P(Z) \not= P(X) + P(Y)$
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