Bag A contains 3 white and 4 red balls,bag B | vedclass

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(D) Let $W_A, W_B, W_C$ be the events of drawing a white ball from bags $A, B, C$ respectively,and $R_A, R_B, R_C$ be the events of drawing a red ball

Vedclass · Accepted Answer

$\frac{286}{693}$

Vedclass · Answer

(D) Let $W_A, W_B, W_C$ be the events of drawing a white ball from bags $A, B, C$ respectively,and $R_A, R_B, R_C$ be the events of drawing a red ball from bags $A, B, C$ respectively.$P(W_A) = \frac{3}{7}, P(R_A) = \frac{4}{7}$$P(W_B) = \frac{4}{9}, P(R_B) = \frac{5}{9}$$P(W_C) = \frac{5}{11}, P(R_C) = \frac{6}{11}$We need one white and two red balls. This can happen in three mutually exclusive cases:Case $I$: White from $A$,Red from $B$,Red from $C$: $P_1 = \frac{3}{7} 	imes \frac{5}{9} 	imes \frac{6}{11} = \frac{90}{693}$Case $II$: Red from $A$,White from $B$,Red from $C$: $P_2 = \frac{4}{7} 	imes \frac{4}{9} 	imes \frac{6}{11} = \frac{96}{693}$Case $III$: Red from $A$,Red from $B$,White from $C$: $P_3 = \frac{4}{7} 	imes \frac{5}{9} 	imes \frac{5}{11} = \frac{100}{693}$Total probability $= P_1 + P_2 + P_3 = \frac{90+96+100}{693} = \frac{286}{693}$.

Bag $A$ contains $3$ white and $4$ red balls,bag $B$ contains $4$ white and $5$ red balls,and bag $C$ contains $5$ white and $6$ red balls. If one ball is drawn at random from each of these three bags,then the probability of getting one white and two red balls is

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