An urn contains three white, six red and four black balls. Two balls are selected at random. What is the probability that one ball is red and other is white, given that they are of different colour ?

Options

A. `(2)/(3)`
B. `(1)/(3)`
C. `(1)/(2)`
D. none of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define the Events
Let:
- Event A: The event that two balls drawn are of different colors.
- Event B: The event that one ball is red and the other is white.

### Step 2: Total Number of Balls
The total number of balls in the urn is:
- White balls = 3
- Red balls = 6
- Black balls = 4

Total = 3 + 6 + 4 = 13 balls.

### Step 3: Calculate the Probability of Event A
To find the probability of event A (two balls of different colors), we need to consider the combinations of colors:
1. One white and one red
2. One white and one black
3. One red and one black

Calculating each case:
1. **One white and one red**:
- Ways to choose 1 white from 3: \( \binom{3}{1} = 3 \)
- Ways to choose 1 red from 6: \( \binom{6}{1} = 6 \)
- Total ways for this case: \( 3 \times 6 = 18 \)

2. **One white and one black**:
- Ways to choose 1 white from 3: \( \binom{3}{1} = 3 \)
- Ways to choose 1 black from 4: \( \binom{4}{1} = 4 \)
- Total ways for this case: \( 3 \times 4 = 12 \)

3. **One red and one black**:
- Ways to choose 1 red from 6: \( \binom{6}{1} = 6 \)
- Ways to choose 1 black from 4: \( \binom{4}{1} = 4 \)
- Total ways for this case: \( 6 \times 4 = 24 \)

Now, add all the cases together to find the total ways to choose two balls of different colors:
\[ \text{Total ways for event A} = 18 + 12 + 24 = 54 \]

### Step 4: Total Ways to Choose Any 2 Balls
The total ways to choose any 2 balls from 13 is given by:
\[ \binom{13}{2} = \frac{13 \times 12}{2} = 78 \]

### Step 5: Calculate Probability of Event A
The probability of event A is:
\[ P(A) = \frac{\text{Number of favorable outcomes for A}}{\text{Total outcomes}} = \frac{54}{78} = \frac{27}{39} = \frac{9}{13} \]

### Step 6: Calculate the Probability of Event B given A
Now, we need to find the probability of event B given event A, which is \( P(B|A) \).

Since event B (one red and one white) is a subset of event A (two balls of different colors), we can calculate the probability directly:
- The number of ways to choose 1 red and 1 white:
\[ \text{Ways for event B} = 3 \times 6 = 18 \]

### Step 7: Calculate \( P(B \cap A) \)
Since event B is already included in event A, we have:
\[ P(B \cap A) = \frac{18}{78} = \frac{3}{13} \]

### Step 8: Calculate \( P(B|A) \)
Using the conditional probability formula:
\[ P(B|A) = \frac{P(B \cap A)}{P(A)} = \frac{\frac{3}{13}}{\frac{9}{13}} = \frac{3}{9} = \frac{1}{3} \]

### Final Answer
Thus, the probability that one ball is red and the other is white, given that they are of different colors, is:
\[ \boxed{\frac{1}{3}} \]