Question

A die is rolled. Consider the following events:
$A = \{1, 2, 5\}$, $B = \{3, 5\}$, $C = \{2, 3, 4, 5\}$
Find the following conditional probabilities:
(i) $P(A|C)$ and $P(C|A)$
(ii) $P(A \cap B|C)$ and $P(A \cup B|C)$}

Hint:

For discrete sample spaces, you can find conditional probability $P(X|Y)$ quickly by ignoring the main sample space $S$ entirely. Treat set $Y$ as your new universe, and count how many elements of $X$ live inside it!

Answer 1

Concept: This problem deals with conditional probability within a finite sample space. When a fair six-sided die is rolled, the complete sample space $S$ is: \[ S = \{1, 2, 3, 4, 5, 6\} \implies n(S) = 6 \] The formula for the conditional probability of an event $X$ given that event $Y$ has occurred is defined as: \[ P(X|Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{n(X \cap Y)}{n(Y)} \] Let us list the elements and counts for each given event set:
• $A = \{1, 2, 5\} \implies n(A) = 3$
• $B = \{3, 5\} \implies n(B) = 2$
• $C = \{2, 3, 4, 5\} \implies n(C) = 4$

Step 1: Finding intersections needed for part (i)
To compute $P(A|C)$ and $P(C|A)$, we first identify the elements common to sets $A$ and $C$: \[ A \cap C = \{1, 2, 5\} \cap \{2, 3, 4, 5\} = \{2, 5\} \implies n(A \cap C) = 2 \]

Step 2: Calculating $P(A|C)$ and $P(C|A)$
Using the conditional probability formulas: \[ P(A|C) = \frac{n(A \cap C)}{n(C)} = \frac{2}{4} = \frac{1}{2} \] \[ P(C|A) = \frac{n(A \cap C)}{n(A)} = \frac{2}{3} \]

Step 3: Finding operations for part (ii)
For the second part, let us determine the compound sets $(A \cap B)$ and $(A \cup B)$ first: \[ A \cap B = \{1, 2, 5\} \cap \{3, 5\} = \{5\} \] \[ A \cup B = \{1, 2, 5\} \cup \{3, 5\} = \{1, 2, 3, 5\} \] Next, find the intersections of these compound sets with condition set $C$:
• For the first case: \[ (A \cap B) \cap C = \{5\} \cap \{2, 3, 4, 5\} = \{5\} \implies n((A \cap B) \cap C) = 1 \]
• For the second case: \[ (A \cup B) \cap C = \{1, 2, 3, 5\} \cap \{2, 3, 4, 5\} = \{2, 3, 5\} \implies n((A \cup B) \cap C) = 3 \]

Step 4: Calculating $P(A \cap B|C)$ and $P(A \cup B|C)$
Using the definition of conditional probability with respect to the reduced sample space of $C$ ($n(C) = 4$): \[ P(A \cap B|C) = \frac{n((A \cap B) \cap C)}{n(C)} = \frac{1}{4} \] \[ P(A \cup B|C) = \frac{n((A \cup B) \cap C)}{n(C)} = \frac{3}{4} \]

Step 5: Summary of Results
Gathering all our calculated values:
• (i) $P(A|C) = \frac{1}{2}$ and $P(C|A) = \frac{2}{3}$
• (ii) $P(A \cap B|C) = \frac{1}{4}$ and $P(A \cup B|C) = \frac{3}{4}$ These values line up with Option (A).