We are given the following probabilities:

• $P(A) = \frac{1}{4}$
• $P(B) = \frac{1}{2}$
• $P(A \cap B) = \frac{1}{8}$

We need to find:

(i) $P(A \cup B)$ (the probability of $A$ or $B$):

The formula for $P(A \cup B)$ is:

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

Substitute the given values:

$P(A \cup B) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8}$

To simplify, find a common denominator (8):

$P(A \cup B) = \frac{2}{8} + \frac{4}{8} - \frac{1}{8} = \frac{5}{8}$

So, $P(A \cup B) = \frac{5}{8}$.

(ii) $P(\neg A \cap \neg B)$ (the probability of not $A$ and not $B$):

The probability of not $A$ and not $B$ is:

$P(\neg A \cap \neg B) = 1 - P(A \cup B)$

We already know that $P(A \cup B) = \frac{5}{8}$, so:

$P(\neg A \cap \neg B) = 1 - \frac{5}{8} = \frac{3}{8}$

Final Answers:

(i) $P(A \cup B) = \frac{5}{8}$

(ii) $P(\neg A \cap \neg B) = \frac{3}{8}$

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Would you like more details or have any questions?

Here are some related questions:

1. What is the relationship between $P(A \cup B)$ and the inclusion-exclusion principle?
2. How would the calculation change if events $A$ and $B$ were mutually exclusive?
3. What is the significance of $P(\neg A \cap \neg B)$ in probability theory?
4. How can you use Venn diagrams to visualize these probabilities?
5. How does the formula for $P(A \cup B)$ extend to three or more events?

Tip: When working with events, always remember to check whether they are independent, mutually exclusive, or conditional, as these properties affect how probabilities are calculated.