Box A contains 2 black and 3 red balls, while | vedclass

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(B) Let $P(B) = p$. Given that $P(A) = 2P(B)$, we have $P(A) = 2p$. Since $P(A) + P(B) = 1$, we get $2p + p = 1$, so $3p = 1$, which means $p = \frac{

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$\frac{10}{31}$

Vedclass · Answer

(B) Let $P(B) = p$. Given that $P(A) = 2P(B)$, we have $P(A) = 2p$. Since $P(A) + P(B) = 1$, we get $2p + p = 1$, so $3p = 1$, which means $p = \frac{1}{3}$. Thus, $P(B) = \frac{1}{3}$ and $P(A) = \frac{2}{3}$.Let $R$ be the event of drawing a red ball.The probability of drawing a red ball from Box $A$ is $P(R|A) = \frac{3}{2+3} = \frac{3}{5}$.The probability of drawing a red ball from Box $B$ is $P(R|B) = \frac{4}{3+4} = \frac{4}{7}$.Using Bayes' Theorem, the probability that the ball came from Box $B$ given that it is red is:$P(B|R) = \frac{P(B) \cdot P(R|B)}{P(A) \cdot P(R|A) + P(B) \cdot P(R|B)}$$P(B|R) = \frac{\frac{1}{3} \cdot \frac{4}{7}}{\frac{2}{3} \cdot \frac{3}{5} + \frac{1}{3} \cdot \frac{4}{7}}$$P(B|R) = \frac{\frac{4}{21}}{\frac{6}{15} + \frac{4}{21}} = \frac{\frac{4}{21}}{\frac{2}{5} + \frac{4}{21}}$$P(B|R) = \frac{\frac{4}{21}}{\frac{42 + 20}{105}} = \frac{\frac{4}{21}}{\frac{62}{105}} = \frac{4}{21} \cdot \frac{105}{62} = \frac{4 \cdot 5}{62} = \frac{20}{62} = \frac{10}{31}$.

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