Let A and B be two events such that `P(AnnB')=0.20,P(A'nnB)=0.15,P(A'nnB')=0.1`thenP(A//B)` is equal to

Options

A. `11//14`
B. `2//11`
C. `2//7`
D. `1//7`

Explanation

To solve the problem, we need to find \( P(A | B) \) given the probabilities:

- \( P(A \cap B') = 0.20 \)
- \( P(A' \cap B) = 0.15 \)
- \( P(A' \cap B') = 0.10 \)

### Step-by-step Solution:

1. **Identify the probabilities:**
- Let \( P(A \cap B) = x \).
- We know:
\[ P(A \cap B') = 0.20 \quad (1) \]
\[ P(A' \cap B) = 0.15 \quad (2) \]
\[ P(A' \cap B') = 0.10 \quad (3) \]

2. **Use the total probability:**
The total probability of all possible outcomes must equal 1:
\[ P(A \cap B) + P(A \cap B') + P(A' \cap B) + P(A' \cap B') = 1 \]
Substituting the known values:
\[ x + 0.20 + 0.15 + 0.10 = 1 \]
Simplifying this gives:
\[ x + 0.45 = 1 \]
Therefore:
\[ x = 1 - 0.45 = 0.55 \quad (4) \]

3. **Find \( P(A) \) and \( P(B) \):**
- Now, we can find \( P(A) \):
\[ P(A) = P(A \cap B) + P(A \cap B') = x + 0.20 = 0.55 + 0.20 = 0.75 \quad (5) \]
- Next, find \( P(B) \):
\[ P(B) = P(A \cap B) + P(A' \cap B) = x + 0.15 = 0.55 + 0.15 = 0.70 \quad (6) \]

4. **Calculate \( P(A | B) \):**
The conditional probability \( P(A | B) \) is given by:
\[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{x}{P(B)} = \frac{0.55}{0.70} \quad (7) \]

5. **Simplify the fraction:**
To simplify \( \frac{0.55}{0.70} \):
\[ \frac{0.55}{0.70} = \frac{55}{70} = \frac{11}{14} \quad (8) \]

6. **Final Answer:**
Thus, the final answer is:
\[ P(A | B) = \frac{11}{14} \]