Key Concepts and Formulas

1. Independent Events: Two events A and B are independent if the occurrence of one does not affect the probability of the other. Mathematically, this is defined as: $P(A \cap B) = P(A) \cdot P(B)$ A crucial property of independent events is that if A and B are independent, then the following pairs of events are also independent:

   • A and B' (complement of B): $P(A \cap B') = P(A) \cdot P(B')$
   • A' (complement of A) and B: $P(A' \cap B) = P(A') \cdot P(B)$
   • A' and B': $P(A' \cap B') = P(A') \cdot P(B')$

2. Conditional Probability: The probability of event X occurring given that event Y has already occurred is defined as: $P(X | Y) = \frac{P(X \cap Y)}{P(Y)}$ provided $P(Y) > 0$.

3. Probability of Union: For any two events A and B: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

4. Probability of Complement: The probability of the complement of an event X is: $P(X') = 1 - P(X)$

Step-by-Step Solution

Given: $P(A) = \frac{1}{3}$ $P(B) = \frac{1}{6}$ A and B are independent events.

Step 1: Calculate $P(A \cap B)$

• What we are doing: Finding the probability of both A and B occurring.
• Why: This is a direct application of the definition of independent events and will be used in subsequent calculations, particularly for the union. Since A and B are independent: $P(A \cap B) = P(A) \cdot P(B)$ Substituting the given values: $P(A \cap B) = \frac{1}{3} \cdot \frac{1}{6} = \frac{1}{18}$

Step 2: Calculate $P(A \cup B)$

• What we are doing: Finding the probability of A or B (or both) occurring.
• Why: This value is required to evaluate Option (A). Using the formula for the probability of union: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ Substituting the given values and the calculated $P(A \cap B)$: $P(A \cup B) = \frac{1}{3} + \frac{1}{6} - \frac{1}{18}$ To add and subtract these fractions, find a common denominator, which is 18: $P(A \cup B) = \frac{6}{18} + \frac{3}{18} - \frac{1}{18} = \frac{6+3-1}{18} = \frac{8}{18} = \frac{4}{9}$

Step 3: Calculate Probabilities of Complements $P(A')$ and $P(B')$

• What we are doing: Finding the probabilities that events A and B do not occur.
• Why: These values will be needed to evaluate Options (C) and (D). Using the formula for the probability of a complement: $P(A') = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3}$ $P(B') = 1 - P(B) = 1 - \frac{1}{6} = \frac{5}{6}$

Step 4: Evaluate Option (A) $P\left( {{A \over {A \cup B}}} \right)$

• What we are doing: Calculating the conditional probability of A given $A \cup B$.
• Why: To check if the statement in Option (A) is true. Using the conditional probability formula: $P(A | A \cup B) = \frac{P(A \cap (A \cup B))}{P(A \cup B)}$ The event $A \cap (A \cup B)$ means elements that are in A AND also in the union of A and B. If an element is in A, it is necessarily in $A \cup B$. Therefore, $A \cap (A \cup B) = A$. So, $P(A \cap (A \cup B)) = P(A) = \frac{1}{3}$. Now, substitute the values: $P(A | A \cup B) = \frac{P(A)}{P(A \cup B)} = \frac{\frac{1}{3}}{\frac{4}{9}} = \frac{1}{3} \cdot \frac{9}{4} = \frac{9}{12} = \frac{3}{4}$ Option (A) states $P\left( {{A \over {A \cup B}}} \right) = {1 \over 4}$. Our calculation yields $\frac{3}{4}$. Thus, Option (A) is FALSE.

Step 5: Evaluate Option (B) $P\left( {{A \over B}} \right)$

• What we are doing: Calculating the conditional probability of A given B.
• Why: To check if the statement in Option (B) is true. Using the conditional probability formula: $P(A | B) = \frac{P(A \cap B)}{P(B)}$ Since A and B are independent, $P(A \cap B) = P(A) \cdot P(B)$. $P(A | B) = \frac{P(A) \cdot P(B)}{P(B)} = P(A)$ Substitute the given value for $P(A)$: $P(A | B) = \frac{1}{3}$ Option (B) states $P\left( {{A \over B}} \right) = {2 \over 3}$. Our calculation yields $\frac{1}{3}$. Thus, Option (B) is FALSE.

Step 6: Evaluate Option (C) $P\left( {{{A'} \over {B'}}} \right)$

• What we are doing: Calculating the conditional probability of A' given B'.
• Why: To check if the statement in Option (C) is true. Using the conditional probability formula: $P(A' | B') = \frac{P(A' \cap B')}{P(B')}$ Since A and B are independent, their complements A' and B' are also independent. Therefore, $P(A' \cap B') = P(A') \cdot P(B')$. $P(A' | B') = \frac{P(A') \cdot P(B')}{P(B')} = P(A')$ Substitute the calculated value for $P(A')$: $P(A' | B') = \frac{2}{3}$ Option (C) states $P\left( {{{A'} \over {B'}}} \right) = {1 \over 3}$. Our calculation yields $\frac{2}{3}$. Thus, Option (C) is FALSE.

Step 7: Evaluate Option (D) $P\left( {{A \over {B'}}} \right)$

• What we are doing: Calculating the conditional probability of A given B'.
• Why: To check if the statement in Option (D) is true. Using the conditional probability formula: $P(A | B') = \frac{P(A \cap B')}{P(B')}$ Since A and B are independent, A and B' are also independent. Therefore, $P(A \cap B') = P(A) \cdot P(B')$. $P(A | B') = \frac{P(A) \cdot P(B')}{P(B')} = P(A)$ Substitute the given value for $P(A)$: $P(A | B') = \frac{1}{3}$ Option (D) states $P\left( {{A \over {B'}}} \right) = {1 \over 3}$. Our calculation yields $\frac{1}{3}$. Thus, Option (D) is TRUE.

Common Mistakes & Tips

• Misunderstanding Independence: A common error is confusing independence with mutual exclusivity. If A and B are independent, $P(A \cap B) = P(A)P(B)$. If A and B are mutually exclusive, $P(A \cap B) = 0$. These are distinct concepts.
• Conditional Probability Identity for Independent Events: Remember that if A and B are independent, then $P(A|B) = P(A)$ and $P(B|A) = P(B)$. This property extends to complements (e.g., $P(A|B') = P(A)$ and $P(A'|B') = P(A')$). This shortcut can save time.
• Set Theory Identities: Simplifying terms like $A \cap (A \cup B) = A$ is crucial for correct calculations. Visualizing with Venn diagrams can be helpful.

Summary

This problem tested our understanding of independent events and conditional probability. We first calculated necessary probabilities such as $P(A \cap B)$, $P(A \cup B)$, $P(A')$, and $P(B')$. Then, we systematically evaluated each option using the conditional probability formula and the properties of independent events. We found that Option (D) was the only statement that held true based on the given probabilities.

The final answer is $\boxed{D}$