Given:

• $P(A) = 0.3$
• $P(B) = 0.4$
• $P(A \cup B) = 0.6$

1. Find $P(A \cap B)$:

We use the formula for the union of two events: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ Substituting the given values: $0.6 = 0.3 + 0.4 - P(A \cap B)$ Solving for $P(A \cap B)$: $P(A \cap B) = 0.3 + 0.4 - 0.6 = 0.1$

Thus, $P(A \cap B) = 0.1$.

2. Find $P(A|B)$:

The conditional probability of $A$ given $B$ is given by: $P(A|B) = \frac{P(A \cap B)}{P(B)}$ Substituting the known values: $P(A|B) = \frac{0.1}{0.4} = 0.25$

Thus, $P(A|B) = 0.25$.

3. Find $P(A|B^c)$:

Here, $B^c$ represents the complement of $B$. The conditional probability of $A$ given $B^c$ is: $P(A|B^c) = \frac{P(A \cap B^c)}{P(B^c)}$ We know: $P(B^c) = 1 - P(B) = 1 - 0.4 = 0.6$ To find $P(A \cap B^c)$, we use: $P(A) = P(A \cap B) + P(A \cap B^c)$ Thus: $0.3 = 0.1 + P(A \cap B^c)$ Solving for $P(A \cap B^c)$: $P(A \cap B^c) = 0.3 - 0.1 = 0.2$ Now, calculate $P(A|B^c)$: $P(A|B^c) = \frac{0.2}{0.6} = \frac{1}{3} \approx 0.3333$

Thus, $P(A|B^c) \approx 0.3333$.

4. Are A and B independent?

Two events $A$ and $B$ are independent if: $P(A \cap B) = P(A) \times P(B)$ We already know $P(A \cap B) = 0.1$, $P(A) = 0.3$, and $P(B) = 0.4$. Now check: $P(A) \times P(B) = 0.3 \times 0.4 = 0.12$ Since $P(A \cap B) = 0.1 \neq 0.12$, the events $A$ and $B$ are not independent.

Summary of Results:

1. $P(A \cap B) = 0.1$
2. $P(A|B) = 0.25$
3. $P(A|B^c) \approx 0.3333$
4. A and B are not independent because $P(A \cap B) \neq P(A) \times P(B)$.

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Would you like further details or clarifications on any part? Here's a list of related questions:

1. How do you compute the probability of the union of two events?
2. What is the general formula for conditional probability?
3. How do you calculate the complement of an event and its probability?
4. Why is it important to test for independence in probability theory?
5. How would you approach a problem if the events were independent?

Tip: When working with conditional probabilities, always remember the definition and how it relates to the joint and marginal probabilities!