1. **Problem statement:** A speaks the truth 3 out of 4 times, B speaks the truth 7 out of 10 times. They agree that a white ball has been drawn from a bag of 6 differently colored balls (only one white ball). Find the probability that the statement "a white ball has been drawn" is true given their agreement. 2. **Define events:** - Let $W$ be the event "white ball drawn". - Let $A$ be the event "A tells the truth" with $P(A) = \frac{3}{4}$. - Let $B$ be the event "B tells the truth" with $P(B) = \frac{7}{10}$. 3. **Prior probability:** Since there is 1 white ball out of 6, $P(W) = \frac{1}{6}$ and $P(\neg W) = \frac{5}{6}$. 4. **Probability both agree the statement is true:** They agree the statement is true means both say "white ball drawn". - If $W$ is true, both tell truth: $P(agree|W) = P(A) \times P(B) = \frac{3}{4} \times \frac{7}{10} = \frac{21}{40}$. - If $W$ is false, both lie (say white ball drawn when it is not): $P(agree|\neg W) = P(\neg A) \times P(\neg B) = \frac{1}{4} \times \frac{3}{10} = \frac{3}{40}$. 5. **Total probability of agreement:** $$P(agree) = P(agree|W)P(W) + P(agree|\neg W)P(\neg W) = \frac{21}{40} \times \frac{1}{6} + \frac{3}{40} \times \frac{5}{6} = \frac{21}{240} + \frac{15}{240} = \frac{36}{240} = \frac{3}{20}.$$ 6. **Use Bayes' theorem to find $P(W|agree)$:** $$P(W|agree) = \frac{P(agree|W)P(W)}{P(agree)} = \frac{\frac{21}{40} \times \frac{1}{6}}{\frac{3}{20}} = \frac{\frac{21}{240}}{\frac{3}{20}} = \frac{21}{240} \times \frac{20}{3} = \frac{21 \times 20}{240 \times 3} = \frac{420}{720} = \frac{7}{12}.$$ **Final answer:** The probability that the statement "a white ball has been drawn" is true given their agreement is $\boxed{\frac{7}{12}}$.