A sample consists of six points S={(I,j)|i=1,2,j=1,2,3}. Suppose P((I,j)) =`k/(i+j)`. Let A = {(I,j)|i+j=4} and B={(I,j)|j=2} , then P(B|A) is equal to

Options

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

Answer

To solve the problem, we need to find \( P(B|A) \), which is the conditional probability of event B given event A. Here’s the step-by-step solution:

### Step 1: Define the Sample Space S
The sample consists of six points defined as:
\[ S = \{(i,j) | i=1,2 \text{ and } j=1,2,3\} \]
The points in S are:
- (1,1)
- (1,2)
- (1,3)
- (2,1)
- (2,2)
- (2,3)

### Step 2: Define Events A and B
- Event A is defined as:
\[ A = \{(i,j) | i+j=4\} \]
The possible pairs that satisfy this condition are:
- (1,3) because \( 1+3=4 \)
- (2,2) because \( 2+2=4 \)
- (3,1) is not possible since \( i \) can only be 1 or 2.

Thus,
\[ A = \{(1,3), (2,2)\} \]

- Event B is defined as:
\[ B = \{(i,j) | j=2\} \]
The possible pairs that satisfy this condition are:
- (1,2)
- (2,2)

Thus,
\[ B = \{(1,2), (2,2)\} \]

### Step 3: Find the Intersection of A and B
Now we need to find the intersection of events A and B:
\[ A \cap B = \{(i,j) | (i,j) \in A \text{ and } (i,j) \in B\} \]
From our definitions:
- A contains (1,3) and (2,2)
- B contains (1,2) and (2,2)

The common point is:
\[ A \cap B = \{(2,2)\} \]

### Step 4: Calculate Probabilities
- The total number of points in sample space \( S \) is 6.
- The number of favorable outcomes for \( A \) is 2 (the points (1,3) and (2,2)).
- The number of favorable outcomes for \( B \cap A \) is 1 (the point (2,2)).

Now we can calculate the probabilities:
- Probability of A:
\[ P(A) = \frac{\text{Number of favorable outcomes for A}}{\text{Total outcomes}} = \frac{2}{6} = \frac{1}{3} \]

- Probability of \( B \cap A \):
\[ P(B \cap A) = \frac{\text{Number of favorable outcomes for } B \cap A}{\text{Total outcomes}} = \frac{1}{6} \]

### Step 5: Calculate Conditional Probability \( P(B|A) \)
Using the formula for conditional probability:
\[ P(B|A) = \frac{P(B \cap A)}{P(A)} \]
Substituting the values we found:
\[ P(B|A) = \frac{\frac{1}{6}}{\frac{2}{6}} = \frac{1}{2} \]

### Final Answer
Thus, the probability \( P(B|A) \) is:
\[ \boxed{\frac{1}{2}} \]