Key Concepts and Formulas

1. Conditional Probability: The probability of an event $C$ occurring given that another event $D$ has already occurred is denoted by $P\left( {{C \over D}} \right)$ (read as "probability of $C$ given $D$"). Its formal definition is: $P\left( {{C \over D}} \right) = \frac{{P\left( {C \cap D} \right)}}{{P\left( D \right)}}$ This formula is valid only when $P(D) \ne 0$, a condition explicitly stated in the problem. The term $C \cap D$ represents the intersection of events $C$ and $D$, meaning both $C$ and $D$ occur simultaneously.

2. Subset of Events ($C \subset D$): When event $C$ is a subset of event $D$, it means that every outcome leading to event $C$ also leads to event $D$. In essence, if $C$ occurs, $D$ must also occur. This relationship has two crucial mathematical implications:

   • The intersection of $C$ and $D$ is $C$ itself: $C \cap D = C$.
   • The probability of $C$ cannot exceed the probability of $D$: $P(C) \le P(D)$.

3. Probability Axioms: For any event $E$, its probability $P(E)$ must satisfy $0 \le P(E) \le 1$. This fundamental property is essential when dealing with inequalities involving probabilities.

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Step-by-Step Solution

Step 1: Apply the Definition of Conditional Probability

Our objective is to determine the relationship involving $P\left( {{C \over D}} \right)$. We begin by writing down its definition, which is the foundational formula for the quantity we need to evaluate. According to the formula for conditional probability: $P\left( {{C \over D}} \right) = \frac{{P\left( {C \cap D} \right)}}{{P\left( D \right)}}$ Why this step? This is the starting point for any problem involving conditional probability. It provides the initial expression that we will manipulate using the given conditions.

Step 2: Utilize the Subset Condition ($C \subset D$)

The problem statement provides a critical piece of information: $C \subset D$. This means that event $C$ is a subset of event $D$.

• Implication of $C \subset D$: If event $C$ occurs, it is guaranteed that event $D$ has also occurred. Therefore, the outcomes common to both $C$ and $D$ (their intersection) are precisely the outcomes that constitute event $C$.
• Mathematical Consequence: $C \cap D = C$. Now, we substitute this simplification into our conditional probability formula from Step 1: $P\left( {{C \over D}} \right) = \frac{{P\left( C \right)}}{{P\left( D \right)}}$ Why this step? This is a crucial simplification. By incorporating the given subset condition, we transform the expression for $P\left( {{C \over D}} \right)$ into a simpler form involving only $P(C)$ and $P(D)$, which sets the stage for comparison.

Step 3: Analyze the Range of $P(D)$

From the basic axioms of probability, we know that for any event $D$, its probability $P(D)$ must lie between 0 and 1, inclusive: $0 \le P(D) \le 1$ The problem statement explicitly provides an additional condition: $P(D) \ne 0$. Combining these two facts, we can refine the possible range for $P(D)$: $0 < P(D) \le 1$ Why this step? Understanding the precise range of $P(D)$ is essential for the next step, where we will compare the fraction $\frac{P(C)}{P(D)}$ with $P(C)$. The properties of division by a number between 0 and 1 (but not including 0) are key to establishing the correct inequality.

Step 4: Establish the Inequality

We have the expression $P\left( {{C \over D}} \right) = \frac{{P\left( C \right)}}{{P\left( D \right)}}$ from Step 2, and we know $0 < P(D) \le 1$ from Step 3. Let's analyze how $\frac{P(C)}{P(D)}$ compares to $P(C)$ based on the value of $P(D)$:

• Case 1: $P(D) = 1$ If $P(D) = 1$, it means $D$ is a sure event (it always occurs). In this specific scenario, our formula becomes: $P\left( {{C \over D}} \right) = \frac{{P\left( C \right)}}{1} = P\left( C \right)$ So, in this case, $P\left( {{C \over D}} \right) = P\left( C \right)$.

• Case 2: $0 < P(D) < 1$ If $P(D)$ is a positive number strictly less than 1 (e.g., $0.5$, $0.25$, $0.9$), then its reciprocal, $\frac{1}{P(D)}$, will be strictly greater than 1. For example, if $P(D) = 0.5$, then $\frac{1}{P(D)} = \frac{1}{0.5} = 2$. Since $P(C)$ is a non-negative probability ($P(C) \ge 0$), multiplying $P(C)$ by a number strictly greater than 1 will result in a value strictly greater than $P(C)$ (unless $P(C)=0$, in which case it remains $0$). Thus, if $0 < P(D) < 1$, then $\frac{1}{P(D)} > 1$. Multiplying both sides by $P(C)$ (which is non-negative, so the inequality direction is preserved): $P(C) \cdot \frac{1}{P(D)} \ge P(C) \cdot 1$ $\frac{P(C)}{P(D)} \ge P(C)$ This implies $P\left( {{C \over D}} \right) \ge P\left( C \right)$.

Combining both cases, whether $P(D)=1$ or $0 < P(D) < 1$, we can definitively conclude that: $P\left( {{C \over D}} \right) \ge P\left( C \right)$ Why this step? This is the final logical step where we synthesize all the information to arrive at the desired inequality. The property that dividing a non-negative number by a value between 0 and 1 (inclusive of 1, exclusive of 0) will either increase or maintain its value is fundamental here.

Step 5: Compare with Options

Our derived result is $P\left( {{C \over D}} \right) \ge P\left( C \right)$. Let's compare this with the given options: (A) $P\left( {{C \over D}} \right) \ge P\left( C \right)$ (B) $P\left( {{C \over D}} \right) < P\left( C \right)$ (C) $P\left( {{C \over D}} \right) = {{P\left( D \right)} \over {P\left( C \right)}}$ (D) $P\left( {{C \over D}} \right) = P\left( C \right)$

Our result matches option (A).

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Common Mistakes & Tips

• Misunderstanding $C \subset D$: A common pitfall is not correctly interpreting $C \subset D$. Remember that it implies $C \cap D = C$ and $P(C) \le P(D)$. Incorrectly assuming $C \cup D = C$ or other relationships can lead to errors.
• Incorrectly Handling Probability Ranges: Always remember that probabilities are numbers between 0 and 1. When you divide by a number between 0 and 1 (exclusive of 0), the result is generally larger than the numerator. Forgetting this can lead to mistakenly choosing option (D) if one only considers $P(D)=1$ or doesn't fully grasp the effect of division by a fraction.
• Intuition for Conditional Probability with Subsets: If $C \subset D$, then for $C$ to occur, $D$ must occur. If we are already given that $D$ has occurred, we have effectively reduced our sample space to just the outcomes in $D$. Within this smaller, restricted sample space $D$, the probability of $C$ occurring (which is all of $C$ within $D$) will typically be higher than its probability in the entire original sample space (unless $D$ itself is the entire sample space, i.e., $P(D)=1$). This intuition supports why $P\left( {{C \over D}} \right)$ is generally greater than or equal to $P(C)$.

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Summary

This problem tests our understanding of conditional probability and the implications of one event being a subset of another. By first applying the definition of conditional probability, $P\left( {{C \over D}} \right) = \frac{{P\left( {C \cap D} \right)}}{{P\left( D \right)}}$, we then utilized the given condition $C \subset D$ to simplify the intersection, leading to $C \cap D = C$. This allowed us to rewrite the conditional probability as $P\left( {{C \over D}} \right) = \frac{{P\left( C \right)}}{{P\left( D \right)}}$. Finally, considering the valid range for $P(D)$ ($0 < P(D) \le 1$), we established that dividing $P(C)$ by a number less than or equal to 1 (but greater than 0) will result in a value greater than or equal to $P(C)$. Therefore, $P\left( {{C \over D}} \right) \ge P\left( C \right)$.

The final answer is $\boxed{A}$