Urn $A$ contains 6 red 4 black balls and urn $B$ contains 4 red and 6 black balls. One ball is drawn

Question

Urn $A$ contains 6 red 4 black balls and urn $B$ contains 4 red and 6 black balls. One ball is drawn at random from urn $A$ and placed in urn $B$. Then one ball is drawn at random from urn $B$ and placed in urn $A$. If one ball is now drawn at random from urn $A$, the probability that it is found to be red is $\frac{m}{n}$ in the lowst form. find m+n

Professor PC Thomas Classes · Answer

Transferring balls from Urn $A$ to Urn $B$ and then $\operatorname{Urn} B$ to $\operatorname{Urn} A$ may be as follows\ (Red, Red) or (Red, Black) or (Black, Red) or (Black, Black)\ $	herefore \quad$ Required probability $$=\frac{6}{10} 	imes \frac{5}{11} 	imes \frac{6}{10}+\frac{4}{10} 	imes \frac{7}{11} 	imes \frac{6}{10}+\frac{6}{10} 	imes \frac{6}{11} 	imes \frac{5}{10} +\frac{4}{10} 	imes \frac{4}{11} 	imes \frac{7}{10}=\frac{64}{110}=\frac{32}{55} $$