Bag I contains 3 red and 4 black balls and Ba | vedclass

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(A) Let $E_{1}$ be the event that a red ball is transferred from Bag $I$ to Bag $II$,and $E_{2}$ be the event that a black ball is transferred from Ba

Vedclass · Accepted Answer

$\frac{16}{31}$

Vedclass · Answer

(A) Let $E_{1}$ be the event that a red ball is transferred from Bag $I$ to Bag $II$,and $E_{2}$ be the event that a black ball is transferred from Bag $I$ to Bag $II$.The probabilities are $P(E_{1}) = \frac{3}{7}$ and $P(E_{2}) = \frac{4}{7}$.Let $A$ be the event that the ball drawn from Bag $II$ is red.If $E_{1}$ occurs,Bag $II$ now contains $5$ red and $5$ black balls. Thus,$P(A|E_{1}) = \frac{5}{10} = \frac{1}{2}$.If $E_{2}$ occurs,Bag $II$ now contains $4$ red and $6$ black balls. Thus,$P(A|E_{2}) = \frac{4}{10} = \frac{2}{5}$.We need to find $P(E_{2}|A)$. By Bayes' Theorem:$P(E_{2}|A) = \frac{P(E_{2})P(A|E_{2})}{P(E_{1})P(A|E_{1}) + P(E_{2})P(A|E_{2})}$$P(E_{2}|A) = \frac{(\frac{4}{7}) 	imes (\frac{2}{5})}{(\frac{3}{7}) 	imes (\frac{1}{2}) + (\frac{4}{7}) 	imes (\frac{2}{5})}$$P(E_{2}|A) = \frac{\frac{8}{35}}{\frac{3}{14} + \frac{8}{35}} = \frac{\frac{8}{35}}{\frac{15+16}{70}} = \frac{8}{35} 	imes \frac{70}{31} = \frac{16}{31}$.

Bag $I$ contains $3$ red and $4$ black balls and Bag $II$ contains $4$ red and $5$ black balls. One ball is transferred from Bag $I$ to Bag $II$ and then a ball is drawn from Bag $II$. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

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