Problem 65
Question
If \(C\) and \(D\) are two events such that \(C \subset D\) and \(P(D) \neq\) 0 , then the correct statement among the following is (A) \(P(C \mid D)=\frac{P(D)}{P(C)}\) (B) \(P(C \mid D)=P(C)\) (C) \(P(C \mid D) \geq P(C)\) (D) \(P(C \mid D)
Step-by-Step Solution
Verified Answer
The correct statement is (D) \(P(C \mid D)
1Step 1: Understand the Problem
We are given two events, \(C\) and \(D\), where \(C \subset D\), meaning all outcomes of \(C\) are also outcomes of \(D\), and \(P(D) eq 0\). We need to determine the correct statement about the conditional probability \(P(C \mid D)\).
2Step 2: Recall Conditional Probability Formula
Conditional probability \(P(C \mid D)\) is defined as \(\frac{P(C \cap D)}{P(D)}\). Since \(C \subset D\), \(C \cap D = C\), so \(P(C \mid D) = \frac{P(C)}{P(D)}\).
3Step 3: Analyze Each Statement
Review the choices and compare them to the conditional probability expression. \(P(C \mid D) = \frac{P(C \cap D)}{P(D)} = \frac{P(C)}{P(D)}\). None of the statements given reflect this formula, so we eliminate option (A). Since \(P(C \mid D) = \frac{P(C)}{P(D)}\), this is smaller than or equal to \(P(C)\), thus ruling out (B) and confirming (D) is correct.
4Step 4: Verify Correctness
We found \(P(C \mid D) = \frac{P(C)}{P(D)}\) is always lesser than or equal to \(P(C)\) because \(P(D)\) is greater than 0, making \(\frac{1}{P(D)} \geq 1\). Therefore, the correct answer is statement (D).
Key Concepts
Understanding Events and SubsetsExploring Probability InequalitiesConditional Probability Formulas Simplified
Understanding Events and Subsets
Imagine two boxes. One is inside the other. This is what it's like when we say that event \(C\) is a subset of event \(D\), written as \(C \subset D\). In this setup, every outcome that falls into event \(C\) also falls into event \(D\).
This is a simple way to visualize the connection between events when one is entirely included in the other, just like one box completely fitting inside a larger box. Hence, if something is true for event \(C\), it must also be true for event \(D\).
This concept plays a critical role in probability because we can predict the relationships between these events. For example, the probability of \(C\) happening is wrapped up in the probability of \(D\), and we can use this relationship to form helpful insights into their likelihoods.
This is a simple way to visualize the connection between events when one is entirely included in the other, just like one box completely fitting inside a larger box. Hence, if something is true for event \(C\), it must also be true for event \(D\).
This concept plays a critical role in probability because we can predict the relationships between these events. For example, the probability of \(C\) happening is wrapped up in the probability of \(D\), and we can use this relationship to form helpful insights into their likelihoods.
Exploring Probability Inequalities
Probability inequalities help us understand how likely one event is compared to another. Because event \(C\) is a subset of event \(D\), we can see that \(P(C) \leq P(D)\).
What does this mean? Essentially, it tells us that the likelihood of event \(C\) can't be greater than the likelihood of event \(D\). Why? Because every outcome in \(C\) is already included in \(D\).
This is a helpful guideline. When calculating probabilities, such inequalities allow us to check if our results make sense. It's a logical tool to understand and confirm the relationships between events, especially when they're nested or dependent upon one another.
What does this mean? Essentially, it tells us that the likelihood of event \(C\) can't be greater than the likelihood of event \(D\). Why? Because every outcome in \(C\) is already included in \(D\).
This is a helpful guideline. When calculating probabilities, such inequalities allow us to check if our results make sense. It's a logical tool to understand and confirm the relationships between events, especially when they're nested or dependent upon one another.
Conditional Probability Formulas Simplified
Conditional probability is like focusing on a smaller piece of a bigger picture. Imagine you know event \(D\) happened, and you want to find the likelihood of \(C\) occurring out of those situations.
The formula \(P(C \mid D) = \frac{P(C \cap D)}{P(D)}\) shows this. Here, \(C \cap D = C\) because \(C\) is a subset of \(D\).
This simplifies things to \(P(C \mid D) = \frac{P(C)}{P(D)}\), turning our focus entirely to the part covered by both \(C\) and \(D\).
The formula \(P(C \mid D) = \frac{P(C \cap D)}{P(D)}\) shows this. Here, \(C \cap D = C\) because \(C\) is a subset of \(D\).
This simplifies things to \(P(C \mid D) = \frac{P(C)}{P(D)}\), turning our focus entirely to the part covered by both \(C\) and \(D\).
- This formula tells us that the probability of \(C\), given \(D\), depends on how \(C\) fits within \(D\).
- The division by \(P(D)\) shrinks the probability of \(C\) relative to how outcomes fit into \(D\).
- It effectively measures a proportion and helps us evaluate the chance of \(C\) when considering only \(D\)'s occurrence.
Other exercises in this chapter
Problem 61
If \(A_{1}, A_{2}, \ldots, A_{n}\) are \(n\) independent events such that \(P(A)\) \(=\frac{1}{i+1}, i=1,2, \ldots, n\). The probability that none of the \(n\)
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Consider 5 independent Bernoulli's trials each with probability of success \(p\). If the probability of at least one failure is greater than or equal to \(31 /
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Three numbers are chosen at random without replacement from \(\\{1,2,3, \ldots, 8]\). The probability that their minimum is 3 , given that their maximum is 6 ,
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A class consists of 80 students, 25 of them are girls. If 10 of the students are rich and 20 of the students are fair complexioned, then the probability of sele
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