The teacher hasn't uploaded a solution for this question yet.
For a single die roll, the sample space is $S = \{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is $n(S) = 6$. The probability of any single outcome is $1/6$.
We use the formula $P(E|F) = \frac{P(E \cap F)}{P(F)}$. Given $A=\{1,2,5\}$ and $C=\{2,3,4,5\}$, the intersection $A \cap C = \{2, 5\}$. Thus, $n(A \cap C) = 2$.
$$P(A|C) = \frac{n(A \cap C)}{n(C)} = \frac{2}{4} = \frac{1}{2}$$
$$P(C|A) = \frac{n(C \cap A)}{n(A)} = \frac{2}{3}$$
First, find $A \cap B = \{5\}$ and $A \cup B = \{1, 2, 3, 5\}$.
For $P(A \cap B | C)$: The intersection $(A \cap B) \cap C = \{5\}$. Thus, $n((A \cap B) \cap C) = 1$.
$$P(A \cap B | C) = \frac{1}{4}$$
For $P(A \cup B | C)$: The intersection $(A \cup B) \cap C = \{2, 3, 5\}$. Thus, $n((A \cup B) \cap C) = 3$.
$$P(A \cup B | C) = \frac{3}{4}$$
Final Answer: (i) P(A|C)=1/2, P(C|A)=2/3; (ii) P(A∩B|C)=1/4, P(A∪B|C)=3/4
AI generated content. Review strictly for academic accuracy.