To solve the problem, we will follow these steps:
### Step 1: Understand the problem
We have two bags:
- Bag A contains: 2 white, 1 black, and 3 red balls.
- Bag B contains: 3 black, 2 red, and n white balls.
We randomly choose one bag and draw 2 balls, which are found to be 1 red and 1 black. We need to find the value of n given that the probability that both balls are from Bag A is \( \frac{6}{11} \).
### Step 2: Define the probabilities
Let:
- \( P(A) = \frac{1}{2} \) (probability of choosing Bag A)
- \( P(B) = \frac{1}{2} \) (probability of choosing Bag B)
We need to calculate the total probability of drawing 1 red and 1 black ball from both bags.
### Step 3: Calculate the probability of drawing 1 red and 1 black from Bag A
From Bag A:
- Total balls = 2 white + 1 black + 3 red = 6 balls.
- The number of ways to choose 1 red and 1 black:
\[
\text{Ways} = \binom{3}{1} \cdot \binom{1}{1} = 3 \cdot 1 = 3
\]
- The total ways to choose any 2 balls from Bag A:
\[
\text{Total ways} = \binom{6}{2} = 15
\]
- Therefore, the probability of drawing 1 red and 1 black from Bag A:
\[
P(\text{1 red, 1 black} | A) = \frac{3}{15} = \frac{1}{5}
\]
### Step 4: Calculate the probability of drawing 1 red and 1 black from Bag B
From Bag B:
- Total balls = 3 black + 2 red + n white = \( n + 5 \) balls.
- The number of ways to choose 1 red and 1 black:
\[
\text{Ways} = \binom{3}{1} \cdot \binom{2}{1} = 3 \cdot 2 = 6
\]
- The total ways to choose any 2 balls from Bag B:
\[
\text{Total ways} = \binom{n + 5}{2} = \frac{(n + 5)(n + 4)}{2}
\]
- Therefore, the probability of drawing 1 red and 1 black from Bag B:
\[
P(\text{1 red, 1 black} | B) = \frac{6}{\frac{(n + 5)(n + 4)}{2}} = \frac{12}{(n + 5)(n + 4)}
\]
### Step 5: Calculate the total probability of drawing 1 red and 1 black
Using the law of total probability:
\[
P(\text{1 red, 1 black}) = P(A) \cdot P(\text{1 red, 1 black} | A) + P(B) \cdot P(\text{1 red, 1 black} | B)
\]
Substituting the values:
\[
P(\text{1 red, 1 black}) = \frac{1}{2} \cdot \frac{1}{5} + \frac{1}{2} \cdot \frac{12}{(n + 5)(n + 4)}
\]
\[
= \frac{1}{10} + \frac{6}{(n + 5)(n + 4)}
\]
### Step 6: Set up the equation using the given probability
We know that:
\[
P(A | \text{1 red, 1 black}) = \frac{P(A) \cdot P(\text{1 red, 1 black} | A)}{P(\text{1 red, 1 black})} = \frac{6}{11}
\]
This gives us:
\[
\frac{\frac{1}{2} \cdot \frac{1}{5}}{P(\text{1 red, 1 black})} = \frac{6}{11}
\]
Substituting \( P(\text{1 red, 1 black}) \):
\[
\frac{\frac{1}{10}}{\frac{1}{10} + \frac{6}{(n + 5)(n + 4)}} = \frac{6}{11}
\]
### Step 7: Cross-multiply to solve for n
Cross-multiplying gives:
\[
11 \cdot \frac{1}{10} = 6 \left( \frac{1}{10} + \frac{6}{(n + 5)(n + 4)} \right)
\]
\[
\frac{11}{10} = \frac{6}{10} + \frac{36}{(n + 5)(n + 4)}
\]
\[
\frac{11}{10} - \frac{6}{10} = \frac{36}{(n + 5)(n + 4)}
\]
\[
\frac{5}{10} = \frac{36}{(n + 5)(n + 4)}
\]
\[
\frac{1}{2} = \frac{36}{(n + 5)(n + 4)}
\]
Cross-multiplying gives:
\[
(n + 5)(n + 4) = 72
\]
Expanding:
\[
n^2 + 9n + 20 = 72
\]
\[
n^2 + 9n - 52 = 0
\]
### Step 8: Solve the quadratic equation
Using the quadratic formula:
\[
n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 1 \cdot (-52)}}{2 \cdot 1}
\]
\[
= \frac{-9 \pm \sqrt{81 + 208}}{2} = \frac{-9 \pm \sqrt{289}}{2} = \frac{-9 \pm 17}{2}
\]
Calculating the two possible values:
1. \( n = \frac{8}{2} = 4 \)
2. \( n = \frac{-26}{2} = -13 \) (not valid)
Thus, the valid solution is:
\[
n = 4
\]
### Final Answer:
The value of n is \( 4 \).
