We are given two independent throws of a fair die, with outcomes $X$ and $Y$. Let $Z = X + Y$ be the sum of the outcomes, and $U = Z \pmod{6}$ be the remainder when $Z$ is divided by 6. We need to check the independence statements.

Statement 1: $X$ and $Z$ are independent

For $X$ and $Z$ to be independent, $P(Z|X) = P(Z)$ must hold for all possible values.

• The possible values for $X$ are $\{1, 2, 3, 4, 5, 6\}$.
• The possible values for $Y$ are $\{1, 2, 3, 4, 5, 6\}$.
• The possible values for $Z$ range from $1+1=2$ to $6+6=12$.
• Let's calculate $P(Z=3)$. $Z=3$ occurs if $(X, Y)$ is $(1, 2)$ or $(2, 1)$. So, $P(Z=3) = P(1,2) + P(2,1) = \frac{1}{36} + \frac{1}{36} = \frac{2}{36} = \frac{1}{18}$.
• Now let's calculate $P(Z=3|X=1)$. If $X=1$, then $Z = 1 + Y$. For $Z=3$, we need $1+Y=3$, which means $Y=2$. Since $X$ and $Y$ are independent, $P(Y=2|X=1) = P(Y=2) = \frac{1}{6}$. Thus, $P(Z=3|X=1) = \frac{1}{6}$.
• Since $P(Z=3|X=1) = \frac{1}{6}$ and $P(Z=3) = \frac{1}{18}$, they are not equal.

Therefore, $X$ and $Z$ are not independent. Statement 1 is false.

Statement 4: $Y$ and $Z$ are not independent

This statement is symmetric to the first one. For $Y$ and $Z$ to be independent, $P(Z|Y) = P(Z)$ must hold.

• We already calculated $P(Z=3) = \frac{1}{18}$.
• Let's calculate $P(Z=3|Y=1)$. If $Y=1$, then $Z = X + 1$. For $Z=3$, we need $X+1=3$, which means $X=2$. Since $X$ and $Y$ are independent, $P(X=2|Y=1) = P(X=2) = \frac{1}{6}$. Thus, $P(Z=3|Y=1) = \frac{1}{6}$.
• Since $P(Z=3|Y=1) = \frac{1}{6}$ and $P(Z=3) = \frac{1}{18}$, they are not equal.

Therefore, $Y$ and $Z$ are not independent. Statement 4 is true.

Statement 3: $Z$ and $U$ are independent

For $Z$ and $U$ to be independent, $P(U|Z) = P(U)$ must hold.

• $U = Z \pmod{6}$. This means that if we know the value of $Z$, the value of $U$ is determined exactly.
• For example, consider $Z=6$. Then $U = 6 \pmod{6} = 0$. So, $P(U=0|Z=6) = 1$.
• Now let's calculate the marginal probability $P(U=0)$. $U=0$ occurs when $Z$ is a multiple of 6. Possible values for $Z$ are $\{2, ..., 12\}$. So, $Z$ can be 6 or 12.
• $P(Z=6) = 5/36$ (pairs: (1,5), (2,4), (3,3), (4,2), (5,1)).
• $P(Z=12) = 1/36$ (pair: (6,6)).
• $P(U=0) = P(Z=6) + P(Z=12) = \frac{5}{36} + \frac{1}{36} = \frac{6}{36} = \frac{1}{6}$.
• Since $P(U=0|Z=6) = 1$ and $P(U=0) = \frac{1}{6}$, they are not equal.

Therefore, $Z$ and $U$ are not independent. Statement 3 is false.

Statement 2: $X$ and $U$ are independent

For $X$ and $U$ to be independent, $P(U|X) = P(U)$ must hold for all possible values.

• First, let's find the distribution of $U$. We found $P(U=0) = 1/6$. By symmetry or calculation, the distribution of $Z \pmod{6}$ for two dice rolls is uniform over $\{0, 1, 2, 3, 4, 5\}$.
• $P(U=0) = P(Z=6) + P(Z=12) = (5+1)/36 = 6/36 = 1/6$.
• $P(U=1) = P(Z=7) = 6/36 = 1/6$.
• $P(U=2) = P(Z=2) + P(Z=8) = (1+5)/36 = 6/36 = 1/6$.
• $P(U=3) = P(Z=3) + P(Z=9) = (2+4)/36 = 6/36 = 1/6$.
• $P(U=4) = P(Z=4) + P(Z=10) = (3+3)/36 = 6/36 = 1/6$.
• $P(U=5) = P(Z=5) + P(Z=11) = (4+2)/36 = 6/36 = 1/6$.
• So, $P(U=k) = 1/6$ for $k \in \{0, 1, 2, 3, 4, 5\}$.
• Now consider $P(U|X=x)$ for any $x \in \{1, ..., 6\}$. $Z = x + Y$. $U = (x+Y) \pmod{6}$.
• As $Y$ takes values $\{1, 2, 3, 4, 5, 6\}$ each with probability $1/6$, the values $x+Y$ take the sequence $\{x+1, x+2, x+3, x+4, x+5, x+6\}$. These are 6 consecutive integers.
• Any set of 6 consecutive integers, when taken modulo 6, results in each remainder $\{0, 1, 2, 3, 4, 5\}$ exactly once. For example, if $x=2$, the sums are $\{3, 4, 5, 6, 7, 8\}$, and modulo 6 they are $\{3, 4, 5, 0, 1, 2\}$.
• Therefore, for any fixed $x$, $P(U=k | X=x) = 1/6$ for each $k \in \{0, 1, 2, 3, 4, 5\}$.
• Since $P(U=k|X=x) = 1/6$ and $P(U=k) = 1/6$, $X$ and $U$ are independent. Statement 2 is true.

The true statements are 2 (B) and 4 (D).