First bag contains 3 red and 5 black balls an | vedclass

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(B) Let $R_1$ and $B_1$ be the events of drawing a red and black ball from the first bag,respectively. $P(R_1) = \frac{3}{8}$,$P(B_1) = \frac{5}{8}$.L

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$\frac{21}{40}$

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(B) Let $R_1$ and $B_1$ be the events of drawing a red and black ball from the first bag,respectively. $P(R_1) = \frac{3}{8}$,$P(B_1) = \frac{5}{8}$.Let $R_2$ and $B_2$ be the events of drawing a red and black ball from the second bag,respectively. $P(R_2) = \frac{6}{10} = \frac{3}{5}$,$P(B_2) = \frac{4}{10} = \frac{2}{5}$.The event that one ball is red and the other is black can happen in two mutually exclusive ways: (Red from first $AND$ Black from second) $OR$ (Black from first $AND$ Red from second).Required Probability $= P(R_1) 	imes P(B_2) + P(B_1) 	imes P(R_2)$.$= (\frac{3}{8} 	imes \frac{4}{10}) + (\frac{5}{8} 	imes \frac{6}{10})$.$= \frac{12}{80} + \frac{30}{80} = \frac{42}{80} = \frac{21}{40}$.

First bag contains $3$ red and $5$ black balls and second bag contains $6$ red and $4$ black balls. $A$ ball is drawn from each bag. The probability that one ball is red and the other is black,is

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