The question asks to show that if $A$ and $B$ are independent attributes, then the attributes:

1. $\alpha$ and $B$ are also independent.
2. $A$ and $\beta$ are also independent.

Explanation:

Let’s denote the probabilities related to the attributes:

• Let $P(A)$ and $P(B)$ be the probabilities of events $A$ and $B$, respectively.
• Independence of $A$ and $B$ implies that: $P(A \cap B) = P(A) \cdot P(B).$

Let $\alpha$ denote the complement of $A$ (i.e., $\alpha = A^c$), and $\beta$ denote the complement of $B$ (i.e., $\beta = B^c$).

(i) Proving that $\alpha$ and $B$ are independent:

We need to show that: $P(\alpha \cap B) = P(\alpha) \cdot P(B).$ Since $\alpha = A^c$, we can rewrite $P(\alpha \cap B)$ as: $P(\alpha \cap B) = P(A^c \cap B) = P(B) - P(A \cap B).$ Given that $P(A \cap B) = P(A) \cdot P(B)$, we get: $P(\alpha \cap B) = P(B) - P(A) \cdot P(B) = P(B) \cdot (1 - P(A)).$ Notice that $1 - P(A) = P(A^c) = P(\alpha)$, thus: $P(\alpha \cap B) = P(\alpha) \cdot P(B).$ This shows that $\alpha$ and $B$ are independent.

(ii) Proving that $A$ and $\beta$ are independent:

We need to show that: $P(A \cap \beta) = P(A) \cdot P(\beta).$ Since $\beta = B^c$, we can rewrite $P(A \cap \beta)$ as: $P(A \cap \beta) = P(A \cap B^c) = P(A) - P(A \cap B).$ Again, using $P(A \cap B) = P(A) \cdot P(B)$, we get: $P(A \cap \beta) = P(A) - P(A) \cdot P(B) = P(A) \cdot (1 - P(B)).$ Since $1 - P(B) = P(B^c) = P(\beta)$, we obtain: $P(A \cap \beta) = P(A) \cdot P(\beta).$ This confirms that $A$ and $\beta$ are independent.

Would you like more details or have any questions?

Related Questions:

1. How does the concept of independence change if we have conditional probabilities?
2. Can independence between multiple events be generalized beyond two events?
3. How does mutual exclusivity differ from independence in probability theory?
4. What happens if $P(A)$ or $P(B)$ equals zero?
5. How can independence be tested empirically with observed data?

Tip:

When dealing with complements in probability, always check if using $1 - P$ can simplify the expression.