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(B) This is a problem of Bayes' theorem.Let $A$ be the event that all $3$ balls are of different colours.Let $E_1, E_2, E_3$ be the events that Box $1

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$\frac{15}{29}$

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(B) This is a problem of Bayes' theorem.Let $A$ be the event that all $3$ balls are of different colours.Let $E_1, E_2, E_3$ be the events that Box $1$,Box $2$,and Box $3$ are chosen respectively.Since one box is chosen at random,$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$.The probability of drawing $3$ balls of different colours from each box is:$P(A|E_1) = \frac{{}^1C_1 	imes {}^2C_1 	imes {}^3C_1}{{}^6C_3} = \frac{1 	imes 2 	imes 3}{20} = \frac{6}{20} = \frac{3}{10}$$P(A|E_2) = \frac{{}^1C_1 	imes {}^1C_1 	imes {}^2C_1}{{}^4C_3} = \frac{1 	imes 1 	imes 2}{4} = \frac{2}{4} = \frac{1}{2}$$P(A|E_3) = \frac{{}^5C_1 	imes {}^4C_1 	imes {}^1C_1}{{}^{10}C_3} = \frac{5 	imes 4 	imes 1}{120} = \frac{20}{120} = \frac{1}{6}$By Bayes' theorem,the probability that the balls came from Box $2$ is:$P(E_2|A) = \frac{P(E_2) \cdot P(A|E_2)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2) + P(E_3) \cdot P(A|E_3)}$$P(E_2|A) = \frac{\frac{1}{3} 	imes \frac{1}{2}}{\frac{1}{3} 	imes \frac{3}{10} + \frac{1}{3} 	imes \frac{1}{2} + \frac{1}{3} 	imes \frac{1}{6}}$$P(E_2|A) = \frac{\frac{1}{6}}{\frac{1}{10} + \frac{1}{6} + \frac{1}{18}} = \frac{\frac{1}{6}}{\frac{9+15+5}{90}} = \frac{\frac{1}{6}}{\frac{29}{90}} = \frac{1}{6} 	imes \frac{90}{29} = \frac{15}{29}$

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