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(B) Let $E_1, E_2, E_3$ be the events of choosing bags $B_1, B_2, B_3$ respectively. The probabilities of choosing the bags are: $P(E_1) = \frac{3}{6}

Vedclass · Accepted Answer

$\frac{5}{14}$

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(B) Let $E_1, E_2, E_3$ be the events of choosing bags $B_1, B_2, B_3$ respectively. The probabilities of choosing the bags are: $P(E_1) = \frac{3}{6} = \frac{1}{2}$ (for outcomes $2, 3, 5$) $P(E_2) = \frac{2}{6} = \frac{1}{3}$ (for outcomes $4, 6$) $P(E_3) = \frac{1}{6}$ (for outcome $1$) Let $G$ be the event of drawing a green ball. The conditional probabilities are: $P(G|E_1) = \frac{3}{2+3+2} = \frac{3}{7}$ $P(G|E_2) = \frac{2}{3+2+2} = \frac{2}{7}$ $P(G|E_3) = \frac{2}{2+2+3} = \frac{2}{7}$ Using the Law of Total Probability: $P(G) = P(E_1)P(G|E_1) + P(E_2)P(G|E_2) + P(E_3)P(G|E_3)$ $P(G) = \left(\frac{3}{6} 	imes \frac{3}{7}ight) + \left(\frac{2}{6} 	imes \frac{2}{7}ight) + \left(\frac{1}{6} 	imes \frac{2}{7}ight)$ $P(G) = \frac{9}{42} + \frac{4}{42} + \frac{2}{42} = \frac{15}{42} = \frac{5}{14}$

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