A bag contains 4 red and 5 black balls. Anoth | vedclass

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(C) Let $B_1$ be the first bag and $B_2$ be the second bag. Bag $B_1$ contains $4$ red and $5$ black balls (Total = $9$). Bag $B_2$ contains $3$ red a

Vedclass · Accepted Answer

$\frac{25}{54}$

Vedclass · Answer

(C) Let $B_1$ be the first bag and $B_2$ be the second bag. Bag $B_1$ contains $4$ red and $5$ black balls (Total = $9$). Bag $B_2$ contains $3$ red and $6$ black balls (Total = $9$). We draw $1$ ball from $B_1$ and $2$ balls from $B_2$. Total balls drawn = $3$. We need $2$ black and $1$ red ball. There are two cases: Case $I$: The ball from $B_1$ is red and the two balls from $B_2$ are black. $P(Case I) = P(R_1) 	imes P(B_2, B_2) = \frac{4}{9} 	imes \frac{\binom{6}{2}}{\binom{9}{2}} = \frac{4}{9} 	imes \frac{15}{36} = \frac{4}{9} 	imes \frac{5}{12} = \frac{20}{108} = \frac{5}{27}$. Case $II$: The ball from $B_1$ is black and the two balls from $B_2$ are one red and one black. $P(Case II) = P(B_1) 	imes P(R_2, B_2) = \frac{5}{9} 	imes \frac{\binom{3}{1} 	imes \binom{6}{1}}{\binom{9}{2}} = \frac{5}{9} 	imes \frac{3 	imes 6}{36} = \frac{5}{9} 	imes \frac{18}{36} = \frac{5}{9} 	imes \frac{1}{2} = \frac{5}{18}$. Total Probability = $P(Case I) + P(Case II) = \frac{5}{27} + \frac{5}{18} = \frac{10 + 15}{54} = \frac{25}{54}$.

$A$ bag contains $4$ red and $5$ black balls. Another bag contains $3$ red and $6$ black balls. If one ball is drawn from the first bag and two balls are drawn from the second bag at random,the probability that out of the three balls,two are black and one is red,is

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