A bag contains 12 red balls 6 white balls. Six balls are drawn one by one without replacement of which at least 4 balls are white. Find the probability that in the next two drawn exactly one white ball is drawn. (Leave the answer in `""^(n)C_(r )`).

Explanation

To solve the problem step by step, we need to find the probability that in the next two drawn balls, exactly one white ball is drawn after having drawn 6 balls from a bag containing 12 red balls and 6 white balls, with the condition that at least 4 of those drawn balls are white.

### Step 1: Define the events
Let:
- \( A \): The event that exactly 4 white balls are drawn in the first 6 draws.
- \( B \): The event that exactly 5 white balls are drawn in the first 6 draws.
- \( C \): The event that all 6 drawn balls are white.

We need to find \( P(E) \), where \( E \) is the event that exactly 1 white ball is drawn in the next 2 draws.

### Step 2: Calculate \( P(E | A) \)
If event \( A \) occurs (4 white balls drawn), then:
- 2 red balls must have been drawn.
- The remaining balls in the bag will be 10 red and 2 white.

The probability of drawing exactly 1 white ball in the next 2 draws can be calculated as follows:
- The number of ways to choose 1 white ball from 2 and 1 red ball from 10:
\[ P(E | A) = \frac{2C1 \cdot 10C1}{12C2} = \frac{2 \cdot 10}{66} = \frac{20}{66} = \frac{10}{33} \]

### Step 3: Calculate \( P(E | B) \)
If event \( B \) occurs (5 white balls drawn), then:
- 1 red ball must have been drawn.
- The remaining balls in the bag will be 11 red and 1 white.

The probability of drawing exactly 1 white ball in the next 2 draws:
- The number of ways to choose 1 white ball from 1 and 1 red ball from 11:
\[ P(E | B) = \frac{1C1 \cdot 11C1}{12C2} = \frac{1 \cdot 11}{66} = \frac{11}{66} = \frac{1}{6} \]

### Step 4: Calculate \( P(E | C) \)
If event \( C \) occurs (6 white balls drawn), then:
- There are no red balls left in the bag.
- The probability of drawing exactly 1 white ball in the next 2 draws is:
\[ P(E | C) = 0 \]

### Step 5: Calculate \( P(A) \), \( P(B) \), and \( P(C) \)
Using combinations:
- Total ways to choose 6 balls from 18:
\[ P(A) = \frac{12C2 \cdot 6C4}{18C6} \]
- For \( B \):
\[ P(B) = \frac{12C1 \cdot 6C5}{18C6} \]
- For \( C \):
\[ P(C) = \frac{6C6}{18C6} = 0 \]

### Step 6: Apply the law of total probability
Now, we can find \( P(E) \):
\[ P(E) = P(E | A) \cdot P(A) + P(E | B) \cdot P(B) + P(E | C) \cdot P(C) \]
Substituting the values:
\[ P(E) = \left(\frac{10}{33} \cdot P(A)\right) + \left(\frac{1}{6} \cdot P(B)\right) + 0 \]

### Step 7: Final Calculation
Now, we can substitute \( P(A) \) and \( P(B) \) into the equation to get the final probability.