Box B1 contains 4 red and 2 white balls; box B2 contains 2 red and 4 white balls. A fair coin is tossed: on heads B1 is chosen, on tails B2. Two…

Within a box the two draws are without replacement. $P(E\mid B_1)=\frac46\cdot\frac35=\frac25$ and $P(E\mid B_2)=\frac26\cdot\frac15=\frac1{15}$. By the total probability theorem $P(E)=\frac12\left(\frac25+\frac1{15}\right)=\frac12\cdot\frac7{15}=\frac7{30}$, so (A) is incorrect ($\frac14$ would come from treating the draws as independent). By Bayes' theorem $P(B_1\mid E)=\frac{\frac12\cdot\frac25}{\frac7{30}}=\frac{6}{7}$, so (B) is incorrect and (C) is correct. For (D), by symmetry each of the events 'first red' and 'second red' has probability $\frac12\left(\frac46+\frac26\right)=\frac12$, whereas the probability that both occur is $\frac7{30}\neq\frac14$, so they are not independent: incorrect. Answer: C only.