1. **State the problem:** We have two urns, each equally likely to be chosen. Urn 1 contains 1 white and 6 red balls, Urn 2 contains 2 white and 5 red balls. A ball is drawn and it is white. We want to find the probability that this white ball came from Urn 2. 2. **Use Bayes' Theorem:** The probability that the white ball came from Urn 2 given that a white ball was drawn is $$P(U2|W) = \frac{P(W|U2)P(U2)}{P(W|U1)P(U1) + P(W|U2)P(U2)}$$ 3. **Calculate each probability:** - Probability of choosing Urn 1 or Urn 2: $$P(U1) = P(U2) = \frac{1}{2}$$ - Probability of drawing white from Urn 1: $$P(W|U1) = \frac{1}{7}$$ (1 white out of 7 total) - Probability of drawing white from Urn 2: $$P(W|U2) = \frac{2}{7}$$ (2 white out of 7 total) 4. **Substitute values into Bayes' formula:** $$P(U2|W) = \frac{\frac{2}{7} \times \frac{1}{2}}{\frac{1}{7} \times \frac{1}{2} + \frac{2}{7} \times \frac{1}{2}}$$ 5. **Simplify numerator and denominator:** $$P(U2|W) = \frac{\frac{2}{14}}{\frac{1}{14} + \frac{2}{14}} = \frac{\frac{2}{14}}{\frac{3}{14}}$$ 6. **Cancel common denominator 14:** $$P(U2|W) = \frac{\cancel{\frac{2}{14}}}{\cancel{\frac{3}{14}}} = \frac{2}{3}$$ 7. **Final answer:** The probability that the white ball came from Urn 2 is $$\boxed{\frac{2}{3}}$$.