This problem deals with conditional probability. Let's analyze and solve it step by step.

Given Information:

1. Three boxes:
   • Box 1: 1 white, 2 red, 3 black balls.
   • Box 2: 2 white, 3 red, 1 black ball.
   • Box 3: 3 white, 1 red, 2 black balls.
2. A box is chosen at random, so the probability of choosing any one box is: $P(\text{Box 1}) = P(\text{Box 2}) = P(\text{Box 3}) = \frac{1}{3}$
3. Two balls are drawn at random, and it is observed that the two balls are 1 red and 1 white.
4. We need to find the probability that the two balls came from Box 2.

Step 1: Total Outcomes for Each Box

Box 1:

• Total balls = $1 + 2 + 3 = 6$.
• Total ways to pick 2 balls = $\binom{6}{2} = 15$.
• Favorable outcomes for 1 red and 1 white:
  • $1$ white ball and $2$ red balls are available.
  • Ways to pick: $\binom{1}{1} \times \binom{2}{1} = 1 \times 2 = 2$
• Probability of drawing 1 red and 1 white from Box 1: $P(1 \text{ red and } 1 \text{ white | Box 1}) = \frac{2}{15}$

Box 2:

• Total balls = $2 + 3 + 1 = 6$.
• Total ways to pick 2 balls = $\binom{6}{2} = 15$.
• Favorable outcomes for 1 red and 1 white:
  • $2$ white balls and $3$ red balls are available.
  • Ways to pick: $\binom{2}{1} \times \binom{3}{1} = 2 \times 3 = 6$
• Probability of drawing 1 red and 1 white from Box 2: $P(1 \text{ red and } 1 \text{ white | Box 2}) = \frac{6}{15}$

Box 3:

• Total balls = $3 + 1 + 2 = 6$.
• Total ways to pick 2 balls = $\binom{6}{2} = 15$.
• Favorable outcomes for 1 red and 1 white:
  • $3$ white balls and $1$ red ball are available.
  • Ways to pick: $\binom{3}{1} \times \binom{1}{1} = 3 \times 1 = 3$
• Probability of drawing 1 red and 1 white from Box 3: $P(1 \text{ red and } 1 \text{ white | Box 3}) = \frac{3}{15}$

Step 2: Using Bayes' Theorem

We want $P(\text{Box 2 | 1 red and 1 white})$. Using Bayes' theorem: $P(\text{Box 2 | 1 red and 1 white}) = \frac{P(1 \text{ red and } 1 \text{ white | Box 2}) \cdot P(\text{Box 2})}{\sum_{i=1}^{3} P(1 \text{ red and } 1 \text{ white | Box } i) \cdot P(\text{Box } i)}$

Numerator:

$P(1 \text{ red and } 1 \text{ white | Box 2}) \cdot P(\text{Box 2}) = \frac{6}{15} \cdot \frac{1}{3} = \frac{6}{45}$

Denominator:

$\sum_{i=1}^{3} P(1 \text{ red and } 1 \text{ white | Box } i) \cdot P(\text{Box } i)$ $= \frac{2}{15} \cdot \frac{1}{3} + \frac{6}{15} \cdot \frac{1}{3} + \frac{3}{15} \cdot \frac{1}{3}$ $= \frac{2}{45} + \frac{6}{45} + \frac{3}{45} = \frac{11}{45}$

Final Probability:

$P(\text{Box 2 | 1 red and 1 white}) = \frac{\frac{6}{45}}{\frac{11}{45}} = \frac{6}{11}$

Final Answer:

The probability that the two balls came from Box 2 is: $\boxed{\frac{6}{11}}$

Do you want more details, or should I clarify any step?

Here are 5 related questions to explore:

1. How would the probability change if the number of balls in Box 2 were different?
2. What happens if we have more than 3 boxes with similar distributions?
3. How does Bayes' theorem generalize to continuous cases in probability?
4. How is the total probability rule used in solving such problems?
5. What if three balls were drawn instead of two?

Tip: Bayes' theorem is especially useful for "reverse probability" problems, where you're given an outcome and need to find the likelihood of a specific cause.