If two events `A` and `B` such that `P(A')=0.3`, `P(B)=0.5` and `P(AnnB)=0.3`, then `P(B//AuuB')` is

Options

A. `3//8`
B. `2//3`
C. `5//6`
D. `1//4`

Explanation

To solve the problem, we need to find \( P(B \mid A \cup B') \) given the following probabilities:

- \( P(A') = 0.3 \)
- \( P(B) = 0.5 \)
- \( P(A \cap B) = 0.3 \)

### Step-by-Step Solution:

**Step 1: Find \( P(A) \)**
We know that \( P(A') = 0.3 \). Therefore,
\[ P(A) = 1 - P(A') = 1 - 0.3 = 0.7 \]

**Step 2: Find \( P(B') \)**
Similarly, since \( P(B) = 0.5 \), we have
\[ P(B') = 1 - P(B) = 1 - 0.5 = 0.5 \]

**Step 3: Find \( P(A \cup B') \)**
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 need to find \( P(A \cap B') \). We can use the relationship:
\[ P(A) = P(A \cap B) + P(A \cap B') \]
Thus,
\[ P(A \cap B') = P(A) - P(A \cap B) = 0.7 - 0.3 = 0.4 \]

Now substituting back into the union formula:
\[ P(A \cup B') = P(A) + P(B') - P(A \cap B') = 0.7 + 0.5 - 0.4 = 0.8 \]

**Step 4: Find \( P(B \cap (A \cup B')) \)**
Using the formula:
\[ P(B \cap (A \cup B')) = P(B \cap A) + P(B \cap B') \]
Since \( P(B \cap A) = P(A \cap B) = 0.3 \) and \( P(B \cap B') = 0 \) (because \( B \) and \( B' \) cannot occur together), we have:
\[ P(B \cap (A \cup B')) = P(A \cap B) + 0 = 0.3 \]

**Step 5: Calculate \( P(B \mid A \cup B') \)**
Now we can use the conditional probability formula:
\[ P(B \mid A \cup B') = \frac{P(B \cap (A \cup B'))}{P(A \cup B')} \]
Substituting the values we found:
\[ P(B \mid A \cup B') = \frac{0.3}{0.8} = \frac{3}{8} \]

### Final Answer:
Thus, the final answer is:
\[ P(B \mid A \cup B') = \frac{3}{8} \]