Let's tackle each part of the problem step by step.

Part (a): Events $A$ and $B$ are independent

(i) $P(A \mid B)$:

For independent events $A$ and $B$, the conditional probability $P(A \mid B)$ is given by:

$P(A \mid B) = P(A)$

Since $P(A) = 0.3$, we have:

$P(A \mid B) = 0.3$

(ii) $P(A \cup B)$:

For independent events, the probability of the union $P(A \cup B)$ is given by:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Since $P(A \cap B) = P(A) \times P(B)$, we have:

$P(A \cup B) = 0.3 + 0.1 - (0.3 \times 0.1)$

$P(A \cup B) = 0.3 + 0.1 - 0.03 = 0.37$

(iii) $P(A \cap B)$:

The probability of the intersection for independent events is:

$P(A \cap B) = P(A) \times P(B)$

$P(A \cap B) = 0.3 \times 0.1 = 0.03$

Part (b): Events $A$ and $B$ are dependent with $P(A \mid B) = 0.4$

(i) $P(B \mid A)$:

Using the conditional probability formula:

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

Given $P(A \mid B) = 0.4$ and $P(B) = 0.1$:

$0.4 = \frac{P(A \cap B)}{0.1}$

So,

$P(A \cap B) = 0.4 \times 0.1 = 0.04$

Now, use the conditional probability formula to find $P(B \mid A)$:

$P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0.04}{0.3} = \frac{4}{30} = \frac{2}{15}$

(ii) $P(A \cap B)$:

We already calculated this in the previous step:

$P(A \cap B) = 0.04$

Summary of Answers

• Part (a):

  • $P(A \mid B) = 0.3$
  • $P(A \cup B) = 0.37$
  • $P(A \cap B) = 0.03$

• Part (b):

  • $P(B \mid A) = \frac{2}{15}$
  • $P(A \cap B) = 0.04$

Would you like any further details or have any questions? Here are some related questions you might find interesting:

1. How do you calculate conditional probability?
2. What is the significance of the independence of events in probability?
3. How does dependence between events affect their probabilities?
4. What are the key differences between union and intersection in probability?
5. How does Bayes' theorem relate to conditional probabilities?

Tip: When working with conditional probabilities, ensure you clearly understand whether events are independent or dependent, as it significantly affects the calculations.