To solve this, we will use the following fundamental probability formulas:

1. Conditional Probability:
   $P(A|B) = \frac{P(A \cap B)}{P(B)} \quad \text{and} \quad P(B|A) = \frac{P(A \cap B)}{P(A)}$
2. Union of Two Events:
   $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

We are given:

• $P(A) = \frac{1}{2}$
• $P(B) = \frac{1}{9}$
• $P(A \cap B) = \frac{1}{18}$

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(i) $P(A | B)$

Using the formula for conditional probability: $P(A | B) = \frac{P(A \cap B)}{P(B)}$

Substitute the given values:
$P(A | B) = \frac{\frac{1}{18}}{\frac{1}{9}}$

Simplify:
$P(A | B) = \frac{1}{18} \times \frac{9}{1} = \frac{9}{18} = \frac{1}{2}$

Thus, $P(A | B) = \frac{1}{2}$.

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(ii) $P(B | A)$

Using the formula for conditional probability: $P(B | A) = \frac{P(A \cap B)}{P(A)}$

Substitute the given values:
$P(B | A) = \frac{\frac{1}{18}}{\frac{1}{2}}$

Simplify:
$P(B | A) = \frac{1}{18} \times \frac{2}{1} = \frac{2}{18} = \frac{1}{9}$

Thus, $P(B | A) = \frac{1}{9}$.

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(iii) $P(A \cup B)$

Using the formula for the union of two events: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Substitute the given values:
$P(A \cup B) = \frac{1}{2} + \frac{1}{9} - \frac{1}{18}$

To simplify, find the least common denominator (LCD), which is 18:
$P(A \cup B) = \frac{9}{18} + \frac{2}{18} - \frac{1}{18}$

Simplify:
$P(A \cup B) = \frac{9 + 2 - 1}{18} = \frac{10}{18} = \frac{5}{9}$

Thus, $P(A \cup B) = \frac{5}{9}$.

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Final Answers:

1. $P(A | B) = \frac{1}{2}$
2. $P(B | A) = \frac{1}{9}$
3. $P(A \cup B) = \frac{5}{9}$

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Do you want further clarification or step-by-step explanations for any part?

Here are 5 related questions you might find useful:

1. How do you interpret conditional probability in real-life scenarios?
2. What is the general formula for the union of three events $P(A \cup B \cup C)$?
3. How do you calculate $P(A | B')$, where $B'$ is the complement of $B$?
4. What is the relationship between $P(A | B)$ and $P(B | A)$?
5. If $A$ and $B$ are independent events, what is the value of $P(A \cap B)$?

Tip: Always check if events $A$ and $B$ are independent, as this simplifies many probability calculations.