Calculating Conditional Probability $P(B | (A \cup \bar{B}))$

This problem requires calculating a conditional probability involving events $A$ and $B$, their complements, unions, and intersections. We are given $P(A)$, $P(B)$, and $P(A \cap \bar{B})$.

Key Given Probabilities

• $P(A) = 0.7$
• $P(B) = 0.4$
• $P(A \cap \bar{B}) = 0.5$

Step 1: Determine $P(A \cap B)$

The probability of event $A$ can be split into two disjoint parts: $A \cap B$ and $A \cap \bar{B}$. Thus, $P(A) = P(A \cap B) + P(A \cap \bar{B})$.

Substituting the given values:

$0.7 = P(A \cap B) + 0.5$

Solving for $P(A \cap B)$:

$P(A \cap B) = 0.7 - 0.5 = 0.2$

Step 2: Determine $P(A \cup \bar{B})$

First, we find the probability of the complement of $B$: $P(\bar{B}) = 1 - P(B)$.

$P(\bar{B}) = 1 - 0.4 = 0.6$

Next, we use the formula for the probability of the union of two events: $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$.

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

Substitute the known probabilities:

$P(A \cup \bar{B}) = 0.7 + 0.6 - 0.5 = 0.8$

Step 3: Determine $P(B \cap (A \cup \bar{B}))$

We need to find the probability of the intersection of event $B$ with the event $(A \cup \bar{B})$.

Using the distributive property of intersection over union: $B \cap (A \cup \bar{B}) = (B \cap A) \cup (B \cap \bar{B})$.

Since $B$ and $\bar{B}$ are complements, their intersection is the empty set: $B \cap \bar{B} = \emptyset$. Therefore, $P(B \cap \bar{B}) = 0$.

The expression simplifies to: $B \cap (A \cup \bar{B}) = B \cap A$.

Thus, $P(B \cap (A \cup \bar{B})) = P(A \cap B)$.

From Step 1, we know $P(A \cap B) = 0.2$. So, $P(B \cap (A \cup \bar{B})) = 0.2$.

Step 4: Calculate $P(B | (A \cup \bar{B}))$

The definition of conditional probability is $P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$.

Applying this formula with $X = B$ and $Y = A \cup \bar{B}$:

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

Substitute the values found in Step 3 and Step 2:

$P(B | (A \cup \bar{B})) = \frac{0.2}{0.8}$

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