The contents of urns I, II and III are as follows: 1 white, 2 black and 3 red balls. 2 white, 1 black and
1 red ball, and 4 white, 5 black and 3 red balls respectively. One urn is chosen at random and two
balls are drawn from it. They happen to be white and red. What is the probability that they come
from urns I

To solve this problem, we can use **Bayes' Theorem**, which is particularly useful for calculating conditional probabilities. Let's break down the problem step by step.

### Step 1: Define the Events
- \( U_1, U_2, U_3 \): The events that Urn I, Urn II, and Urn III are chosen, respectively.
- \( A \): The event that two balls drawn are one white and one red.

### Step 2: Find the Probability of Each Urn Being Chosen
Since one urn is chosen at random, the probability of choosing each urn is:
\[
P(U_1) = P(U_2) = P(U_3) = \frac{1}{3}
\]

### Step 3: Calculate the Probability of Drawing One White and One Red Ball from Each Urn
We calculate \( P(A|U_i) \) for each urn:

1. **Urn I**:
   - Total balls: \( 1 \text{ white} + 2 \text{ black} + 3 \text{ red} = 6 \) balls
   - Probability of drawing 1 white and 1 red:
     \[
     P(A|U_1) = \frac{\text{(Number of white balls)} \times \text{(Number of red balls)}}{\text{Total pairs of balls}} = \frac{1 \times 3}{\binom{6}{2}} = \frac{3}{15} = \frac{1}{5}
     \]

2. **Urn II**:
   - Total balls: \( 2 \text{ white} + 1 \text{ black} + 1 \text{ red} = 4 \) balls
   - Probability of drawing 1 white and 1 red:
     \[
     P(A|U_2) = \frac{2 \times 1}{\binom{4}{2}} = \frac{2}{6} = \frac{1}{3}
     \]

3. **Urn III**:
   - Total balls: \( 4 \text{ white} + 5 \text{ black} + 3 \text{ red} = 12 \) balls
   - Probability of drawing 1 white and 1 red:
     \[
     P(A|U_3) = \frac{4 \times 3}{\binom{12}{2}} = \frac{12}{66} = \frac{2}{11}
     \]

### Step 4: Apply Bayes' Theorem
We want to find the probability that the balls were drawn from Urn I given that they are white and red:
\[
P(U_1|A) = \frac{P(A|U_1) \cdot P(U_1)}{P(A|U_1) \cdot P(U_1) + P(A|U_2) \cdot P(U_2) + P(A|U_3) \cdot P(U_3)}
\]

Substituting the values:
\[
P(U_1|A) = \frac{\frac{1}{5} \times \frac{1}{3}}{\frac{1}{5} \times \frac{1}{3} + \frac{1}{3} \times \frac{1}{3} + \frac{2}{11} \times \frac{1}{3}}
\]

Simplifying:
\[
P(U_1|A) = \frac{\frac{1}{15}}{\frac{1}{15} + \frac{1}{9} + \frac{2}{33}}
\]

Converting all terms to a common denominator:
\[
P(U_1|A) = \frac{\frac{1}{15}}{\frac{22 + 35 + 10}{165}} = \frac{\frac{1}{15}}{\frac{67}{165}} = \frac{11}{67}
\]

### Conclusion
The probability that the balls come from Urn I is \( \frac{11}{67} \).

Would you like more details on any step, or do you have any questions?

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**Further Questions to Explore:**
1. What is Bayes' Theorem and how is it used in probability?
2. How do you calculate the probability of drawing a specific combination of balls from an urn?
3. What is the significance of the common denominator in probability calculations?
4. How would the probability change if different numbers of balls were drawn?
5. What would happen if we had more urns with different distributions of balls?
6. How can the binomial coefficient be used in problems like this?
7. What if the balls were drawn with replacement? How would that affect the probabilities?
8. How does conditional probability differ from simple probability?

**Tip:** Always double-check the total number of outcomes when calculating probabilities, especially in conditional probability problems.

Learn how to solve a probability problem involving drawing balls from urns using Bayes' Theorem. Understand the concept of conditional probability and apply it to determine the probability of drawing specific colored balls from different urns. Suitable for advanced high school students and anyone interested in probability theory.

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