Probability Concepts and Conditional Probability

The question asks us to identify statements that are necessarily true regarding conditional probabilities of two events, A and B, in a discrete probability space. We are given that the probabilities of both events are positive, i.e., $\mathbb{P}(A) > 0$ and $\mathbb{P}(B) > 0$.

Let's recall the definition of conditional probability:

• The conditional probability of event A occurring given that event B has occurred is $\mathbb{P}(A \mid B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$, provided $\mathbb{P}(B) > 0$.
• The conditional probability of event B occurring given that event A has occurred is $\mathbb{P}(B \mid A) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(A)}$, provided $\mathbb{P}(A) > 0$.

Since we are given $\mathbb{P}(A) > 0$ and $\mathbb{P}(B) > 0$, these definitions are always valid for the given events.

Analyzing Each Statement

Statement 1: If $\mathbb{P}(A \mid B) = 0$ then $\mathbb{P}(B \mid A) = 0$.

Let's assume the condition $\mathbb{P}(A \mid B) = 0$ is true. Using the definition of conditional probability:

$\mathbb{P}(A \mid B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} = 0$

Since we are given $\mathbb{P}(B) > 0$, the only way this equation can hold is if the numerator is zero.

$\mathbb{P}(A \cap B) = 0$

Now let's consider the conclusion $\mathbb{P}(B \mid A) = 0$. Using the definition:

$\mathbb{P}(B \mid A) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(A)}$

We found that $\mathbb{P}(A \cap B) = 0$. Substituting this into the expression for $\mathbb{P}(B \mid A)$:

$\mathbb{P}(B \mid A) = \frac{0}{\mathbb{P}(A)}$

Since we are given $\mathbb{P}(A) > 0$, division by $\mathbb{P}(A)$ is allowed, and the result is 0.

$\mathbb{P}(B \mid A) = 0$

Thus, if $\mathbb{P}(A \mid B) = 0$, it necessarily follows that $\mathbb{P}(A \cap B) = 0$, which in turn necessarily implies $\mathbb{P}(B \mid A) = 0$. So, statement 1 is necessarily true.

Statement 2: If $\mathbb{P}(A \mid B) = 1$ then $\mathbb{P}(B \mid A) = 1$.

Let's assume the condition $\mathbb{P}(A \mid B) = 1$ is true. Using the definition:

$\mathbb{P}(A \mid B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} = 1$

Since $\mathbb{P}(B) > 0$, this means $\mathbb{P}(A \cap B) = \mathbb{P}(B)$. This equality implies that event B must be a subset of event A ($B \subseteq A$), because the intersection of A and B is B itself.

Now let's consider the conclusion $\mathbb{P}(B \mid A) = 1$. Using the definition:

$\mathbb{P}(B \mid A) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(A)}$

Since $\mathbb{P}(A \cap B) = \mathbb{P}(B)$, we have:

$\mathbb{P}(B \mid A) = \frac{\mathbb{P}(B)}{\mathbb{P}(A)}$

For $\mathbb{P}(B \mid A)$ to be 1, it must be the case that $\mathbb{P}(B) = \mathbb{P}(A)$. However, knowing that $B \subseteq A$ and $\mathbb{P}(B) > 0, \mathbb{P}(A) > 0$ does not necessarily mean $\mathbb{P}(B) = \mathbb{P}(A)$. For example, consider a sample space $\Omega = \{1, 2, 3\}$ with probabilities $\mathbb{P}(\{1\}) = 0.2$, $\mathbb{P}(\{2\}) = 0.3$, $\mathbb{P}(\{3\}) = 0.5$. Let $A = \{1, 2\}$ and $B = \{1\}$.

• $\mathbb{P}(A) = 0.2 + 0.3 = 0.5$
• $\mathbb{P}(B) = 0.2$
• $\mathbb{P}(A \cap B) = \mathbb{P}(\{1\}) = 0.2$

Check the condition: $\mathbb{P}(A \mid B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} = \frac{0.2}{0.2} = 1$. The condition is met.

Check the conclusion: $\mathbb{P}(B \mid A) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(A)} = \frac{0.2}{0.5} = 0.4$. The conclusion $\mathbb{P}(B \mid A) = 1$ is not met, as $0.4 \neq 1$.

Therefore, statement 2 is not necessarily true.

Statement 3: If $\mathbb{P}(A \mid B) > \mathbb{P}(A)$ then $\mathbb{P}(B \mid A) > \mathbb{P}(B)$.

Let's analyze the condition $\mathbb{P}(A \mid B) > \mathbb{P}(A)$. Using the definition:

$\frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} > \mathbb{P}(A)$

Since $\mathbb{P}(B) > 0$, we can multiply both sides by $\mathbb{P}(B)$ without changing the inequality direction:

$\mathbb{P}(A \cap B) > \mathbb{P}(A) \mathbb{P}(B)$

Now let's analyze the conclusion $\mathbb{P}(B \mid A) > \mathbb{P}(B)$. Using the definition:

$\frac{\mathbb{P}(A \cap B)}{\mathbb{P}(A)} > \mathbb{P}(B)$

Since $\mathbb{P}(A) > 0$, we can multiply both sides by $\mathbb{P}(A)$ without changing the inequality direction:

$\mathbb{P}(A \cap B) > \mathbb{P}(A) \mathbb{P}(B)$

We see that the condition $\mathbb{P}(A \mid B) > \mathbb{P}(A)$ is equivalent to the inequality $\mathbb{P}(A \cap B) > \mathbb{P}(A) \mathbb{P}(B)$. Similarly, the conclusion $\mathbb{P}(B \mid A) > \mathbb{P}(B)$ is also equivalent to $\mathbb{P}(A \cap B) > \mathbb{P}(A) \mathbb{P}(B)$. Therefore, if the condition is true, the inequality $\mathbb{P}(A \cap B) > \mathbb{P}(A) \mathbb{P}(B)$ holds, which in turn makes the conclusion true. So, statement 3 is necessarily true.

This relationship indicates that if event A is more likely given event B than it is overall, then event B is also more likely given event A than it is overall. This describes a positive correlation between the two events.

Statement 4: If $\mathbb{P}(A \mid B) > \mathbb{P}(B)$ then $\mathbb{P}(B \mid A) > \mathbb{P}(A)$.

Let's analyze the condition $\mathbb{P}(A \mid B) > \mathbb{P}(B)$. This is equivalent to $\frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} > \mathbb{P}(B)$, which simplifies to $\mathbb{P}(A \cap B) > (\mathbb{P}(B))^2$ since $\mathbb{P}(B) > 0$.

Now let's analyze the conclusion $\mathbb{P}(B \mid A) > \mathbb{P}(A)$. This is equivalent to $\frac{\mathbb{P}(A \cap B)}{\mathbb{P}(A)} > \mathbb{P}(A)$, which simplifies to $\mathbb{P}(A \cap B) > (\mathbb{P}(A))^2$ since $\mathbb{P}(A) > 0$.

So, the statement claims: If $\mathbb{P}(A \cap B) > (\mathbb{P}(B))^2$, then $\mathbb{P}(A \cap B) > (\mathbb{P}(A))^2$. This is not necessarily true. We can construct a counterexample where $\mathbb{P}(A \cap B)$ is greater than $\mathbb{P}(B)^2$ but not greater than $\mathbb{P}(A)^2$. This would likely occur if $\mathbb{P}(A)$ is much larger than $\mathbb{P}(B)$.

Consider a sample space $\Omega = \{1, 2, 3\}$. Let probabilities be $\mathbb{P}(\{1\}) = 0.05$, $\mathbb{P}(\{2\}) = 0.9$, $\mathbb{P}(\{3\}) = 0.05$.

Let $A = \{1, 2\}$ and $B = \{1\}$.

• $\mathbb{P}(A) = \mathbb{P}(\{1\}) + \mathbb{P}(\{2\}) = 0.05 + 0.9 = 0.95$
• $\mathbb{P}(B) = \mathbb{P}(\{1\}) = 0.05$
• $\mathbb{P}(A \cap B) = \mathbb{P}(\{1\}) = 0.05$

Check the condition: $\mathbb{P}(A \mid B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} = \frac{0.05}{0.05} = 1$. Is $\mathbb{P}(A \mid B) > \mathbb{P}(B)$? Is $1 > 0.05$? Yes, the condition is met.

Check the conclusion: $\mathbb{P}(B \mid A) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(A)} = \frac{0.05}{0.95} = \frac{5}{95} = \frac{1}{19} \approx 0.0526$. Is $\mathbb{P}(B \mid A) > \mathbb{P}(A)$? Is $\frac{1}{19} > 0.95$? No, $0.0526$ is not greater than $0.95$.

Therefore, statement 4 is not necessarily true.

Summary of Findings

Based on our analysis:

• Statement 1 is necessarily true.
• Statement 2 is not necessarily true.
• Statement 3 is necessarily true.
• Statement 4 is not necessarily true.

The statements that are necessarily true are Statement 1 and Statement 3.