If two events A and B are such that `P(A')=0.3,P(B)=0.4and P(AnnB')=0.5,` then find the value of `P[B//AuuB')].`

Explanation

To solve the problem, we need to find the value of \( P(B | A \cup B') \) given the probabilities \( P(A') = 0.3 \), \( P(B) = 0.4 \), and \( P(A \cap B') = 0.5 \).

### Step 1: Calculate \( P(A) \) and \( P(B') \)

Since \( P(A') = 0.3 \), we can find \( P(A) \):
\[ P(A) = 1 - P(A') = 1 - 0.3 = 0.7 \]

Next, we calculate \( P(B') \):
\[ P(B') = 1 - P(B) = 1 - 0.4 = 0.6 \]

### Step 2: Calculate \( P(A \cup B') \)

Using the formula for the probability of the union of two events:
\[ P(A \cup B') = P(A) + P(B') - P(A \cap B') \]
We already have \( P(A) = 0.7 \), \( P(B') = 0.6 \), and \( P(A \cap B') = 0.5 \). Now we can substitute these values:
\[ P(A \cup B') = 0.7 + 0.6 - 0.5 = 0.8 \]

### Step 3: Calculate \( P(B \cap (A \cup B')) \)

Using the formula for the intersection of two events:
\[ P(B \cap (A \cup B')) = P(B \cap A) + P(B \cap B') \]

We know that:
\[ P(B \cap A) = P(B) - P(B \cap A') \quad \text{(since } A' \text{ and } A \text{ are complements)} \]
We need to find \( P(B \cap A') \). We can use the fact that:
\[ P(A \cap B') + P(A' \cap B) + P(A \cap B) + P(A' \cap B') = 1 \]
From the given information, we have:
\[ P(A \cap B') = 0.5, \quad P(B) = 0.4 \quad \text{(which includes } P(A \cap B) + P(A' \cap B)\text{)} \]
Thus:
\[ P(A' \cap B) = P(B) - P(A \cap B) = 0.4 - P(A \cap B) \]
Now we need to find \( P(A \cap B) \). We can use the total probability:
\[ P(A \cap B) + P(A' \cap B) + P(A \cap B') + P(A' \cap B') = 1 \]
Substituting known values:
\[ P(A \cap B) + (0.4 - P(A \cap B)) + 0.5 + P(A' \cap B') = 1 \]
This simplifies to:
\[ 0.5 + P(A' \cap B') = 1 \implies P(A' \cap B') = 0.5 \]
Thus, we can find \( P(A' \cap B) \):
\[ P(A' \cap B) = 0.4 - P(A \cap B) \]
Now we can find \( P(B \cap (A \cup B')) \):
\[ P(B \cap (A \cup B')) = P(B) = 0.4 \]

### Step 4: Calculate \( P(B | A \cup B') \)

Now we can use the formula for conditional probability:
\[ P(B | A \cup B') = \frac{P(B \cap (A \cup B'))}{P(A \cup B')} \]
Substituting the values we found:
\[ P(B | A \cup B') = \frac{0.4}{0.8} = 0.5 \]

### Final Answer
\[ P(B | A \cup B') = 0.5 \]