A binary code $(X, Y, Z)$ is chosen from $\{0, 1\}^3$. Since each digit can be 0 or 1, there are $2^3 = 8$ possible, equally likely outcomes:
$$S = \{(0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0), (1,1,1)\}$$
Let's identify the outcomes for each event and calculate their probabilities:
Event $A$ ($X = Y$): The first two digits match.
$A = \{(0,0,0), (0,0,1), (1,1,0), (1,1,1)\}$ $\implies P(A) = \frac{4}{8} = \frac{1}{2}$
Event $B$ ($Y = Z$): The last two digits match.
$B = \{(0,0,0), (1,0,0), (0,1,1), (1,1,1)\}$ $\implies P(B) = \frac{4}{8} = \frac{1}{2}$
Event $C$ ($X = Z$): The first and last digits match.
$C = \{(0,0,0), (0,1,0), (1,0,1), (1,1,1)\}$ $\implies P(C) = \frac{4}{8} = \frac{1}{2}$
We also need the intersection of these events. If $A$ ($X=Y$) and $B$ ($Y=Z$) both occur, then $X=Y=Z$.
Intersection ($A \cap B$): All three digits match.
$A \cap B = \{(0,0,0), (1,1,1)\}$ $\implies P(A \cap B) = \frac{2}{8} = \frac{1}{4}$
By the same logic, $P(A \cap C) = \frac{1}{4}$ and $P(B \cap C) = \frac{1}{4}$.
The intersection of all three ($A \cap B \cap C$) is simply the event where $X=Y=Z$, so $P(A \cap B \cap C) = \frac{1}{4}$.
Now we will evaluate the Statements :
Statement A: $A$ and $B$ are independent.
Two events are independent if $P(A \cap B) = P(A)P(B)$.
$P(A \cap B) = \frac{1}{4}$
$P(A) \cdot P(B) = \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right) = \frac{1}{4}$
Since $\frac{1}{4} = \frac{1}{4}$, this statement is True.
Statement B: $A$ and $B$ are conditionally independent given $C$.
This requires checking if $P(A \cap B | C) = P(A | C) \cdot P(B | C)$.
$P(A | C) = \frac{P(A \cap C)}{P(C)} = \frac{1/4}{1/2} = \frac{1}{2}$
$P(B | C) = \frac{P(B \cap C)}{P(C)} = \frac{1/4}{1/2} = \frac{1}{2}$
$P(A \cap B | C) = \frac{P(A \cap B \cap C)}{P(C)} = \frac{1/4}{1/2} = \frac{1}{2}$
Now let's check the condition:
$\frac{1}{2} \neq \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right)$
Since $\frac{1}{2} \neq \frac{1}{4}$, they are not conditionally independent. This statement is False.
Statement C: $A, B$, and $C$ are mutually independent.
For three events to be mutually independent, $P(A \cap B \cap C)$ must equal $P(A)P(B)P(C)$.
Since $\frac{1}{4} \neq \frac{1}{8}$, this statement is False.
Statement D: $P(A \cup B | C^c) = 1$.
Event $C^c$ is the complement of $C$, meaning $X \neq Z$. Because our code is binary (only 0s and 1s), if the first and third digits are different, the middle digit ($Y$) must match one of them.
If $Y$ matches $X$, then event $A$ happens.
If $Y$ matches $Z$, then event $B$ happens.
Therefore, if $C^c$ occurs, it is guaranteed that either $A$ or $B$ (or both) will occur. Because it is a certainty, the probability of $A \cup B$ given $C^c$ is $1$. This statement is True.
