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(B) Let $A$ be the first bag and $B$ be the second bag.For bag $A$: $P(	ext{red}) = \frac{3}{8}$,$P(	ext{black}) = \frac{5}{8}$.For bag $B$: $P(	ex

Vedclass · Accepted Answer

$\frac{21}{40}$

Vedclass · Answer

(B) Let $A$ be the first bag and $B$ be the second bag.For bag $A$: $P(	ext{red}) = \frac{3}{8}$,$P(	ext{black}) = \frac{5}{8}$.For bag $B$: $P(	ext{red}) = \frac{6}{10} = \frac{3}{5}$,$P(	ext{black}) = \frac{4}{10} = \frac{2}{5}$.The event that one ball is red and the other is black can happen in two ways:$1$. Red from bag $A$ and black from bag $B$.$2$. Black from bag $A$ and red from bag $B$.Required probability $= (P(	ext{red}_A) 	imes P(	ext{black}_B)) + (P(	ext{black}_A) 	imes P(	ext{red}_B))$$= (\frac{3}{8} 	imes \frac{4}{10}) + (\frac{5}{8} 	imes \frac{6}{10})$$= \frac{12}{80} + \frac{30}{80} = \frac{42}{80} = \frac{21}{40}$.

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

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