Conditional Probability: Identify The Incorrect Statement

Hey guys! Let's dive into a probability problem adapted from ANPEC. We need to figure out which of the following statements about conditional probability is wrong. Conditional probability can be tricky, so let's break each option down.

Analyzing the Statements

Let's go through each of the given options, one by one, to figure out which one isn't playing by the rules of probability. Understanding each statement deeply is super important, and will help solidify our understanding.

(A) 0 ≤ P(A∣B) ≤ 1

This statement is all about the range of conditional probability. It states that the conditional probability of event A given event B, written as P(A∣B), always falls between 0 and 1, inclusive. In simpler terms, it's saying that a conditional probability can never be negative or greater than 1, which makes sense because probabilities represent the likelihood of an event occurring, and this likelihood can only range from 0% to 100%. Think of it like a percentage: you can't have -10% chance or 150% chance of something happening! This is a fundamental property of probability, ensuring that our calculations and interpretations remain within a logical and interpretable framework.

So, why is this true? Well, conditional probability is defined as P(A∣B) = P(A ∩ B) / P(B), provided P(B) > 0. Since both P(A ∩ B) and P(B) are probabilities themselves, they must lie between 0 and 1. Dividing a number between 0 and 1 by another number between 0 and 1 will always result in a value between 0 and 1. This stems directly from the axioms of probability and how conditional probability is derived. Therefore, this statement is correct.

(B) P(A∣B)/P(B∣A) = P(A)/P(B)

This option deals with the relationship between two conditional probabilities and their respective unconditional probabilities. The equation P(A∣B)/P(B∣A) = P(A)/P(B) suggests a proportional relationship between how A affects B and vice versa, linking it to the probabilities of A and B occurring independently. Let’s break this down using the definitions of conditional probability.

We know that P(A∣B) = P(A ∩ B) / P(B) and P(B∣A) = P(A ∩ B) / P(A). If we divide P(A∣B) by P(B∣A), we get: [P(A ∩ B) / P(B)] / [P(A ∩ B) / P(A)] = P(A) / P(B). This is because the P(A ∩ B) terms cancel out. This shows that the relationship holds true mathematically, derived directly from the definitions of conditional probability. It reveals an elegant symmetry: the ratio of how A influences B to how B influences A is simply the ratio of their individual probabilities. Therefore, statement B is correct.

(C) P(A∣B) = P(A∩B)

This statement proposes that the conditional probability of A given B is equal to the joint probability of A and B occurring together. This is a tricky one! The claim P(A∣B) = P(A∩B) is generally NOT true. Conditional probability, P(A∣B), means the probability of A happening given that B has already happened. The joint probability, P(A∩B), is the probability of both A and B happening together. These are only equal under specific circumstances.

Recall that P(A∣B) = P(A ∩ B) / P(B). For P(A∣B) to equal P(A ∩ B), P(B) would have to be equal to 1. In other words, B would have to be a certain event. This is a very specific condition and not generally true for all events A and B. Therefore, this statement is incorrect!

(D) Se P(A) = 0,4, P(B) = 0,8 e P(A∣B) = 0,2, então P(B∣A) = 0,4

This option provides specific probability values and asks us to calculate another conditional probability. Given P(A) = 0.4, P(B) = 0.8, and P(A∣B) = 0.2, we need to verify if P(B∣A) is indeed 0.4. We can use the formulas for conditional probability to check if this holds true.

First, we know P(A∣B) = P(A ∩ B) / P(B). We can rearrange this to find P(A ∩ B): P(A ∩ B) = P(A∣B) * P(B) = 0.2 * 0.8 = 0.16. Now, we can use this value to calculate P(B∣A): P(B∣A) = P(A ∩ B) / P(A) = 0.16 / 0.4 = 0.4. Therefore, this statement is correct.

(E) Se P(B) = 0,6 e P(A∣B) = 0,2, então P(Ac ∪ B) = ?

This option requires us to compute the probability of the union of the complement of A (Ac) and B, given the probabilities of B and A given B. We're given P(B) = 0.6 and P(A∣B) = 0.2, and our task is to find P(Ac ∪ B). To solve this, we can use the properties of probability and set theory.

First, we can express P(Ac ∪ B) using the formula: P(Ac ∪ B) = P(Ac) + P(B) - P(Ac ∩ B). We know that P(Ac) = 1 - P(A). Also, we know P(A∣B) = P(A ∩ B) / P(B), so P(A ∩ B) = P(A∣B) * P(B) = 0.2 * 0.6 = 0.12. Now, we can use the fact that P(B) = P(A ∩ B) + P(Ac ∩ B), so P(Ac ∩ B) = P(B) - P(A ∩ B) = 0.6 - 0.12 = 0.48. Next, we need to find P(A). We know P(A ∩ B) = 0.12, but we don't have enough information to directly find P(A). However, we can express P(Ac ∪ B) = P(Ac) + P(B) - P(Ac ∩ B) = (1 - P(A)) + P(B) - P(Ac ∩ B). Substituting the values, we get P(Ac ∪ B) = (1 - P(A)) + 0.6 - 0.48 = 1 - P(A) + 0.12. Alternatively, we can use the formula P(Ac ∪ B) = P(Ac) + P(B) - P(Ac ∩ B). We have P(Ac ∩ B) = P(B) - P(A ∩ B) = 0.6 - 0.12 = 0.48. Then, P(Ac ∪ B) = (1 - P(A)) + 0.6 - 0.48 = 0.6 + (1 - P(A)) - 0.48 = 1 - P(A) + 0.12. However, since we don't know P(A), let's approach differently:

Let's try expressing ${P(A^c \cup B)}$ as ${P(U) - P(A \cap B^c)}$. ${P(U) = 1}$ and ${P(A \cap B^c) = P(A) - P(A \cap B)}$. Thus, ${P(A^c \cup B) = 1 - (P(A) - P(A \cap B))}$. We know that ${P(A \cap B) = P(A|B) * P(B) = 0.2 * 0.6 = 0.12}$. Therefore, ${P(A^c \cup B) = 1 - P(A) + 0.12}$. However, there seems to be missing information to determine P(A), meaning we cannot compute the exact value, but the expression is mathematically sound given what's presented.

Conclusion

So, after carefully analyzing each statement, we've pinpointed that (C) P(A∣B) = P(A∩B) is the incorrect one. Remember folks, conditional probability isn't the same as joint probability unless the given event is certain! Keep practicing, and you'll nail these concepts in no time!