The problem consists of two parts. Part 1: Given $P(\overline{A}) = 0.3$, $P(B) = 0.4$, and $P(A \cap \overline{B}) = 0.5$, find the conditional probability $P(B | A \cup \overline{B})$. Part 2: Given $P(A) = \frac{1}{4}$, $P(B|A) = \frac{1}{3}$, and $P(A|B) = \frac{1}{2}$, find $P(A \cup B)$.

1. Problem Description

The problem consists of two parts.

Part 1: Given $P(\overline{A}) = 0.3$, $P(B) = 0.4$, and $P(A \cap \overline{B}) = 0.5$, find the conditional probability $P(B | A \cup \overline{B})$.

Part 2: Given $P(A) = \frac{1}{4}$, $P(B|A) = \frac{1}{3}$, and $P(A|B) = \frac{1}{2}$, find $P(A \cup B)$.

2. Solution Steps

Part 1:

We want to find $P(B | A \cup \overline{B})$.

Using the definition of conditional probability:

$P(B | A \cup \overline{B}) = \frac{P(B \cap (A \cup \overline{B}))}{P(A \cup \overline{B})}$.

First, let's find $P(B \cap (A \cup \overline{B}))$.

$B \cap (A \cup \overline{B}) = (B \cap A) \cup (B \cap \overline{B}) = (B \cap A) \cup \emptyset = B \cap A$.

So, $P(B \cap (A \cup \overline{B})) = P(A \cap B)$.

Next, we need to find $P(A \cup \overline{B})$.

Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, we have

$P(A \cup \overline{B}) = P(A) + P(\overline{B}) - P(A \cap \overline{B})$.

We are given $P(\overline{A}) = 0.3$, so $P(A) = 1 - P(\overline{A}) = 1 - 0.3 = 0.7$.

Also, $P(B) = 0.4$, so $P(\overline{B}) = 1 - P(B) = 1 - 0.4 = 0.6$.

Then, $P(A \cup \overline{B}) = 0.7 + 0.6 - 0.5 = 0.8$.

To find $P(A \cap B)$, we use the fact that $P(A) = P(A \cap B) + P(A \cap \overline{B})$.

$0.7 = P(A \cap B) + 0.5$, so $P(A \cap B) = 0.7 - 0.5 = 0.2$.

Therefore, $P(B | A \cup \overline{B}) = \frac{P(A \cap B)}{P(A \cup \overline{B})} = \frac{0.2}{0.8} = \frac{1}{4} = 0.25$.

Part 2:

We want to find $P(A \cup B)$.

Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

We are given $P(A) = \frac{1}{4}$, $P(B|A) = \frac{1}{3}$, and $P(A|B) = \frac{1}{2}$.

We know that $P(B|A) = \frac{P(A \cap B)}{P(A)}$ and $P(A|B) = \frac{P(A \cap B)}{P(B)}$.

So, $\frac{1}{3} = \frac{P(A \cap B)}{\frac{1}{4}}$, which means $P(A \cap B) = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$.

Also, $\frac{1}{2} = \frac{P(A \cap B)}{P(B)}$, which means $P(B) = 2 \cdot P(A \cap B) = 2 \cdot \frac{1}{12} = \frac{1}{6}$.

Therefore, $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{4} + \frac{1}{6} - \frac{1}{12} = \frac{3}{12} + \frac{2}{12} - \frac{1}{12} = \frac{4}{12} = \frac{1}{3}$.

3. Final Answer

Part 1: $P(B | A \cup \overline{B}) = 0.25$

Part 2: $P(A \cup B) = \frac{1}{3}$