For k=1, 2, 3, the box B_k contains k red bal | vedclass

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(B) Let $R$ be the event of drawing a red ball. The contents of the boxes are:$B_1: 1R, 2W \implies P(R|B_1) = \frac{1}{3}$$B_2: 2R, 3W \implies P(R|B

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$\frac{14}{39}$

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(B) Let $R$ be the event of drawing a red ball. The contents of the boxes are:$B_1: 1R, 2W \implies P(R|B_1) = \frac{1}{3}$$B_2: 2R, 3W \implies P(R|B_2) = \frac{2}{5}$$B_3: 3R, 4W \implies P(R|B_3) = \frac{3}{7}$Using Bayes' Theorem, the probability that the ball came from box $B_2$ given that it is red is:$P(B_2|R) = \frac{P(B_2)P(R|B_2)}{P(B_1)P(R|B_1) + P(B_2)P(R|B_2) + P(B_3)P(R|B_3)}$Substituting the values:$P(B_2|R) = \frac{\frac{1}{3} 	imes \frac{2}{5}}{\frac{1}{2} 	imes \frac{1}{3} + \frac{1}{3} 	imes \frac{2}{5} + \frac{1}{6} 	imes \frac{3}{7}}$$P(B_2|R) = \frac{\frac{2}{15}}{\frac{1}{6} + \frac{2}{15} + \frac{1}{14}}$Finding a common denominator for the denominator $(210)$:$P(B_2|R) = \frac{\frac{2}{15}}{\frac{35 + 28 + 15}{210}} = \frac{2}{15} 	imes \frac{210}{78} = \frac{280}{1170} = \frac{14}{39}$

For $k=1, 2, 3$, the box $B_k$ contains $k$ red balls and $(k+1)$ white balls. Let $P(B_1)=\frac{1}{2}$, $P(B_2)=\frac{1}{3}$, and $P(B_3)=\frac{1}{6}$. $A$ box is selected at random and a ball is drawn from it. If a red ball is drawn, then the probability that it has come from box $B_2$ is:

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