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 (1 white, 5 non-white). 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" with $P(W) = \frac{1}{6}$. - Let $\overline{W}$ be "non-white ball drawn" with $P(\overline{W}) = \frac{5}{6}$. - Let $A_T$ be "A tells the truth" with $P(A_T) = \frac{3}{4}$. - Let $B_T$ be "B tells the truth" with $P(B_T) = \frac{7}{10}$. 3. **Calculate probability they agree:** They agree if both say "white" or both say "not white". - Probability both say "white": - If white drawn, both tell truth: $P(W) \times P(A_T) \times P(B_T) = \frac{1}{6} \times \frac{3}{4} \times \frac{7}{10} = \frac{21}{240} = \frac{7}{80}$. - If non-white drawn, both lie (say white): $P(\overline{W}) \times (1 - P(A_T)) \times (1 - P(B_T)) = \frac{5}{6} \times \frac{1}{4} \times \frac{3}{10} = \frac{15}{240} = \frac{1}{16}$. - Probability both say "not white": - If white drawn, both lie (say not white): $P(W) \times (1 - P(A_T)) \times (1 - P(B_T)) = \frac{1}{6} \times \frac{1}{4} \times \frac{3}{10} = \frac{3}{240} = \frac{1}{80}$. - If non-white drawn, both tell truth: $P(\overline{W}) \times P(A_T) \times P(B_T) = \frac{5}{6} \times \frac{3}{4} \times \frac{7}{10} = \frac{105}{240} = \frac{7}{16}$. 4. **Total probability they agree:** $$P(agree) = \frac{7}{80} + \frac{1}{16} + \frac{1}{80} + \frac{7}{16} = \frac{7}{80} + \frac{5}{80} + \frac{1}{80} + \frac{35}{80} = \frac{48}{80} = \frac{3}{5}.$$ 5. **Probability they agree and white is drawn:** $$P(agree \cap W) = P(W) \times P(A_T) \times P(B_T) + P(W) \times (1 - P(A_T)) \times (1 - P(B_T)) = \frac{7}{80} + \frac{1}{80} = \frac{8}{80} = \frac{1}{10}.$$ 6. **Use Bayes' theorem to find probability white ball drawn given agreement:** $$P(W|agree) = \frac{P(agree \cap W)}{P(agree)} = \frac{\frac{1}{10}}{\frac{3}{5}} = \frac{1}{10} \times \frac{5}{3} = \frac{1}{6}.$$ **Final answer:** The probability that the statement "a white ball has been drawn" is true given that A and B agree is $\boxed{\frac{1}{6}}$.