I need to calculate $\Pr(A)$ in the following cases:
- $A = B \cap C \cap D$ and $B, C, D$ are independent events
- $A = (B \cup C) \cap (D \cup C)$ and $B, C, D, E$ are independent events
- $A = (B \cap C) \cup (D \cap C)$ with $(B \cap C)\cap (D \cap C) = \varnothing$ and B, C, D, E are independent events'}
- $A = (B \cap C) \cup (D \cap C)$ and $(B \cap C)\cap (D \cap C) \neq \varnothing$ and $B, C, D, E$ are independent events.''}
- $A = B \cup C \cup D$ und $B, C, D$ are NOT NESCESSARILY independent events.
Progress so far:
Here are some solutions I came up with. I realize that the notation is a little untidy/the simplifications have been inconsistently applied. I would greatly appreciate any feedback.
Task 1: $$ \Pr(A) = \Pr(B) \cdot \Pr(C) \cdot \Pr(D) $$
Task 2: \begin{align*} \Pr(A) &= \big(\Pr(B) + \Pr(C) - \Pr(B \cap C) \big) \cap \big( \Pr(D) + \Pr(C) - \Pr(C \cap D)\big) \\ &= \big(\Pr(B) + \Pr(C) - \Pr(B \cap C) \big) \cdot \big( \Pr(D) + \Pr(C) - \Pr(C \cap D)\big) \end{align*}
Task 3: \begin{align*} \Pr(A) &= \Pr(B \cap C) + \Pr(D \cap C) + \Pr\big( (B \cap C) \cap (D \cap C) \big) \\ &= \Pr(B \cap C) + \Pr(D \cap C) + 0 \\ &= \Pr(B \cap C) + \Pr(D \cap C) \\ &= \Pr(B) \cdot \Pr(C) + \Pr(D) \cdot \Pr(C) \\ &= \Pr(C) \cdot \big( \Pr(B) + \Pr(D) \big) \end{align*}
Task 4: \begin{align*} \Pr(A) &= \Pr(B \cap C) + \Pr(D \cap C) + \Pr\big( (B \cap C) \cap (D \cap C) \big) \\ &= \Pr(C) \cdot \big( \Pr(B) + \Pr(D) \big) + \Pr\big( (B \cap C) \cap (D \cap C) \big) \\ &= \Pr(C) \cdot \big( \Pr(B) + \Pr(D) \big) + \big( (\Pr(B) \cdot \Pr(C)) \cdot (\Pr(D) \cdot \Pr(C) \big) \\ &= \Pr(C) \cdot \big( \Pr(B) + \Pr(D) \big) + \Pr(C)^2 \cdot \Pr(B) \cdot \Pr(D) \\ &= \Pr(C) \cdot \big( \Pr(B) + \Pr(D) + \Pr(C) \cdot \Pr(B) \cdot \Pr(D) \big) \end{align*}
Task 5: $$ \Pr(A) = \Pr(B) + \Pr(C) + \Pr(D) - \Pr(B \cap C) - \Pr(B \cap D) - \Pr(C \cap D) + \Pr(B \cap C \cap D)$$