We have two bags such that the first bag contains $ 1 $ black balls and $ 2 $ red balls and the second bag contains $ 5 $ black balls and $ 1 $ red balls. We throw an unbalanced coin such that the probability of getting a "PILLE" is equal to one third of the probability of getting a "FACE". If we get the back " PILLE" we pull a ball from the second bag and if we get "FACE" we pull a ball from the first bag. The process stops when we pull a red ball. What is the probability of stopping the experiment ?
Attempts: Probability of stopping the experiment is the probility of pulling red ball , I have tried to use conditional probability such that $P(A|B)=\dfrac{p(A\cap B)}{p(B)}$, "B" is the probability of throwing the coin " , "$A|B $ is the probability when the experiment stop" , Now we have $p(B)=1/6$ because we have equiprobaility if we pull any ball from any above bags which is $\frac12$, now my problem is what is $p(A\cap B)$? Events here are dependents according to this exercices this means that we can't have $p(A|B)=p(B) $.? Any help ?
There is no need to apply that formula here as we have not been given an event that has occurred with certainty.
Second, the question does not state if we continue to draw and if so with replacement or without. If we have to find it for the first draw, here is how we can do -
Probability that we will pick a ball from bag $1$ is $3$ times of that we would pick from bag $2$ depending on the biased coin toss.
So probability that we would pick from bag $1$ is $\frac{3}{4}$. Probability that we would pick from bag $2$ is $\frac{1}{4}$. You know the number of balls in each bag to calculate the probability.
The probability of stopping the experiment after first draw = $\frac{3}{4} \times P(R|1) + \frac{1}{4} \times P(R|2)$.
$P(R|1)$ is probability of picking a red ball from a draw from bag $1$ and $P(R|2)$ is probability of picking a red ball from a draw from bag $2$.