If `P(A)= 1//3, P(B) = 1//4 and P(A cap B) = 1//5," then : " P(B'//A')=`

Options

A. `37//40`
B. `37//45`
C. `23//40`
D. `23//45`

Answer

To solve the problem, we need to find \( P(B'|A') \) given that \( P(A) = \frac{1}{3} \), \( P(B) = \frac{1}{4} \), and \( P(A \cap B) = \frac{1}{5} \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the conditional probability \( P(B'|A') \). By the definition of conditional probability, we have:
\[ P(B'|A') = \frac{P(B' \cap A')}{P(A')} \]

2. **Finding \( P(A') \)**:
We know that:
\[ P(A') = 1 - P(A) \]
Substituting the value of \( P(A) \):
\[ P(A') = 1 - \frac{1}{3} = \frac{2}{3} \]

3. **Finding \( P(A \cup B) \)**:
We can use the formula for the union of two events:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the given values:
\[ P(A \cup B) = \frac{1}{3} + \frac{1}{4} - \frac{1}{5} \]
To compute this, we need a common denominator. The least common multiple (LCM) of 3, 4, and 5 is 60. Converting each fraction:
\[ P(A) = \frac{1}{3} = \frac{20}{60}, \quad P(B) = \frac{1}{4} = \frac{15}{60}, \quad P(A \cap B) = \frac{1}{5} = \frac{12}{60} \]
Now substituting these values:
\[ P(A \cup B) = \frac{20}{60} + \frac{15}{60} - \frac{12}{60} = \frac{23}{60} \]

4. **Finding \( P(A' \cap B') \)**:
We can use the complement rule:
\[ P(A' \cap B') = 1 - P(A \cup B) \]
Substituting the value we found:
\[ P(A' \cap B') = 1 - \frac{23}{60} = \frac{37}{60} \]

5. **Finding \( P(B'|A') \)**:
Now we can substitute back into the conditional probability formula:
\[ P(B'|A') = \frac{P(B' \cap A')}{P(A')} = \frac{P(A' \cap B')}{P(A')} \]
Substituting the values we calculated:
\[ P(B'|A') = \frac{\frac{37}{60}}{\frac{2}{3}} = \frac{37}{60} \times \frac{3}{2} \]
Simplifying this:
\[ P(B'|A') = \frac{37 \times 3}{60 \times 2} = \frac{111}{120} = \frac{37}{40} \]

### Final Answer:
Thus, the probability \( P(B'|A') = \frac{37}{40} \).