A bag labeled $A$ contains $4$ red balls and $7$ green balls.
Another bag $B$ contains $6$ red and $5$ green balls.
A ball is transferred from bag $A$ to bag $B$, after which a ball is drawn from $B$.
Find the probability that it is a red ball?
To be honest I have no idea how to approach the question, I assume that there would be $12$ balls in the bag $B$ when it has been transferred. I'm lost.
Best Answer
Hint:
$$P\left(E\right)=P\left(E\mid R\right)P\left(R\right)+P\left(E\mid G\right)P\left(G\right)$$
Here $R$ is the event that the transferred ball is red, $G$ is the event that the transferred ball is green and $E$ denotes the event that the ball taken out bag $B$ is red.
Second hint: $P\left(E\mid R\right)=\frac7{12}$, why?...