Question

Urn $A$ contains $6$ red and $4$ 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 $4k/55$.Find the value of $k$

Answer 1

$R_1:$ ' a red ball is drawn from urn $A$ and placed in $B$'
$B_1:$ ' a black ball is drawn from urn $A$ and placed in $B$'
$R_2:$ ' a black ball is drawn from urn $B$ and placed in $A$'
$B_2:$ 'a red ball is drawn from urn $B$ and placed in $A$'
$R:$ 'a red ball is drawn in the second attempt from $A$'
Then the required probability
$P\left(R_1{R}_{2}R\right)+P\left(R_1{B}_{2}R\right)+P\left(B_1{R}_{2}R\right)+P\left(B_1{B}_{2}R\right)=P\left(R_1\right)P\left(R_2\right)P\left(R\right)+P\left(R_1\right)P\left(B_2\right)P\left(R\right)+P\left(B_1\right)P\left(R_2\right)P\left(R\right)+P\left(B_1\right)P\left(B_2\right)P\left(R\right)$
$=\frac{6}{10}\times \frac{5}{11}\times \frac{6}{10}+\frac{6}{10}\times \frac{6}{11}\times \frac{6}{10}+\frac{4}{10}\times \frac{5}{11}\times \frac{6}{10}+\frac{4}{10}\times \frac{6}{11}\times \frac{6}{10}=\frac{32}{55}\therefore k=8$