There are 3 urns A, B,and C. Urn A contains 4 | vedclass

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Question

(A) Let $R_A, R_B, R_C$ be the events of drawing a red ball from urns $A, B, C$ respectively,and $B_A, B_B, B_C$ be the events of drawing a black ball

Vedclass · Accepted Answer

$\frac{17}{42}$

Vedclass · Answer

(A) Let $R_A, R_B, R_C$ be the events of drawing a red ball from urns $A, B, C$ respectively,and $B_A, B_B, B_C$ be the events of drawing a black ball from urns $A, B, C$ respectively.Probabilities are:$P(R_A) = \frac{4}{7}, P(B_A) = \frac{3}{7}$$P(R_B) = \frac{5}{9}, P(B_B) = \frac{4}{9}$$P(R_C) = \frac{4}{8} = \frac{1}{2}, P(B_C) = \frac{4}{8} = \frac{1}{2}$To get $2$ red balls and $1$ black ball,the possible cases are:$1$. $(R_A, R_B, B_C) = \frac{4}{7} 	imes \frac{5}{9} 	imes \frac{1}{2} = \frac{20}{126}$$2$. $(R_A, B_B, R_C) = \frac{4}{7} 	imes \frac{4}{9} 	imes \frac{1}{2} = \frac{16}{126}$$3$. $(B_A, R_B, R_C) = \frac{3}{7} 	imes \frac{5}{9} 	imes \frac{1}{2} = \frac{15}{126}$Total probability $= \frac{20+16+15}{126} = \frac{51}{126} = \frac{17}{42}$.

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