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

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(B) Let $R$ be the event that a red ball is drawn. We are given the probabilities of selecting each box: $P(B_1) = \frac{1}{2}$,$P(B_2) = \frac{1}{3}$

Vedclass · Accepted Answer

$\frac{14}{39}$

Vedclass · Answer

(B) Let $R$ be the event that a red ball is drawn. We are given the probabilities of selecting each box: $P(B_1) = \frac{1}{2}$,$P(B_2) = \frac{1}{3}$,$P(B_3) = \frac{1}{6}$.The probability of drawing a red ball from each box is:$P(R|B_1) = \frac{1}{1+2} = \frac{1}{3}$$P(R|B_2) = \frac{2}{2+3} = \frac{2}{5}$$P(R|B_3) = \frac{3}{3+4} = \frac{3}{7}$Using the Law of Total Probability,the probability of drawing a red ball is:$P(R) = P(B_1)P(R|B_1) + P(B_2)P(R|B_2) + P(B_3)P(R|B_3)$$P(R) = \left(\frac{1}{2} 	imes \frac{1}{3}ight) + \left(\frac{1}{3} 	imes \frac{2}{5}ight) + \left(\frac{1}{6} 	imes \frac{3}{7}ight)$$P(R) = \frac{1}{6} + \frac{2}{15} + \frac{1}{14} = \frac{35 + 28 + 15}{210} = \frac{78}{210} = \frac{13}{35}$Using Bayes' Theorem,the probability that the ball came from box $B_2$ given that it is red is:$P(B_2|R) = \frac{P(B_2)P(R|B_2)}{P(R)} = \frac{\frac{1}{3} 	imes \frac{2}{5}}{\frac{13}{35}} = \frac{2}{15} 	imes \frac{35}{13} = \frac{2 	imes 7}{3 	imes 13} = \frac{14}{39}$.

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 comes from box $B_2$ is:

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