There are 5 red and 7 white balls in a bag. Two balls are drawn one by one without replacement. Find the probability that first ball is red and second is white.

Explanation

To solve the problem of finding the probability that the first ball drawn is red and the second ball drawn is white, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Total Number of Balls:**
- There are 5 red balls and 7 white balls in the bag.
- Total number of balls = 5 (red) + 7 (white) = 12 balls.

2. **Determine the Probability of Drawing a Red Ball First:**
- The probability of drawing a red ball first (Event A) is given by the ratio of the number of red balls to the total number of balls.
- \( P(A) = \frac{\text{Number of Red Balls}}{\text{Total Number of Balls}} = \frac{5}{12} \).

3. **Update the Total After Drawing the First Ball:**
- After drawing one red ball, the total number of balls left in the bag is 11 (12 - 1).
- The number of white balls remains the same, which is 7.

4. **Determine the Probability of Drawing a White Ball Second:**
- The probability of drawing a white ball second (Event B given A) is given by the ratio of the number of white balls remaining to the total number of balls remaining.
- \( P(B|A) = \frac{\text{Number of White Balls}}{\text{Total Remaining Balls}} = \frac{7}{11} \).

5. **Calculate the Combined Probability:**
- The required probability that the first ball is red and the second ball is white is given by the product of the probabilities of the two events:
- \( P(A \cap B) = P(A) \times P(B|A) = \frac{5}{12} \times \frac{7}{11} \).

6. **Perform the Multiplication:**
- \( P(A \cap B) = \frac{5 \times 7}{12 \times 11} = \frac{35}{132} \).

### Final Answer:
The probability that the first ball drawn is red and the second ball drawn is white is \( \frac{35}{132} \).