Bag P contains 6 red and 4 blue balls and bag Q contains 5 red and...

Let $E_1, E_2 and A$ be the events defined as follows:
$\mathrm{E}_1=$ red ball is transferred from bag $\mathrm{P}$ to bag $\mathrm{Q}$
$\mathrm{E}_2=$ blue ball is transferred from bag $\mathrm{P}$ to bag $\mathrm{Q}$
$\mathrm{A}=$ the ball drawn from bag $\mathrm{Q}$ is blue
As the bag $\mathrm{P}$ contains 6 red and 4 blue balls,
$P\left(E_1\right)=\frac{6}{10}=\frac{3}{5} \text{ and } P\left(E_2\right)=\frac{4}{10}=\frac{2}{5}$
Note that $E_1 and E_2$ are mutually exclusive and exhaustive events.
When $E_1$ has occurred i.e., a red ball has already been transferred from bag $P$ to $Q$, then bag $Q$ will contain 6 red and 6 blue balls, So, $P\left(A \mid E_1\right)=\frac{6}{12}=\frac{1}{2}$ \\ When $E_2$ has occurred i.e., a blue ball has already been transferred from bag $P$ to $Q$, then bag $Q$ will contain 5 red and 7 blue balls, So, $P\left(A \mid E_2\right)=\frac{7}{12}$
By using law of total probability, we get
$P(A)=P \left(E_1\right) P \left(A \mid E_1\right)+P\left(E_2\right)$P \left(A \mid E_2\right)$<br />=$\frac{3}{5}$\times$\frac{1}{2}+\frac{2}{5}$\times$\frac{7}{12}=\frac{8}{15}