Suppose we have two bags of marbles, $C$ and $\bar C$. Marbles are red and green.
Initially, the bags are $C = \{ G, G, R, R, R, R \}$ and $\bar C = \{ G, R \}$. Then one marble is randomly moved from bag $C$ into bag $\bar C$.
Now you draw a marble from $\bar C$
First, what is the probability it's green?
Second, we've found out it is green. Knowing this, what's the probability that we moved a green marble from bag $C$ into bag $\bar C$ earlier?
This problem is different than other conditional probability problems (marble moves from one bag to another). Could someone walk me through this? I'm not sure how to deal with the changing variable.
Best Answer
Let the probability of moving green from $C$ be $P(G_C)=1/3$, then $P(R_C)=2/3$.
Probability of getting a green when $G_C$ has happened, $P(G_{\bar C}|G_C)=2/3$ and similarly $P(G_{\bar C}|R_C)=1/3$.
Since, $P(A|B)=\frac{P(A\cap B)}{P(B)}$, so, $P(G_{\bar C}\cap G_C)=\frac23\frac13$ and $P(G_{\bar C}\cap R_C)=\frac13\frac23$
Using Bayes' Theorem,
$P(G_{\bar C})=\frac{P(G_{\bar C}\cap G_C)+P(G_{\bar C}\cap R_C)}{P(G_C)+P(R_C)}=\frac49$
Here, we have to find $P(G_{\bar C}|G_C)$.
Again using Bayes' Theorem:
$P(G_{\bar C}|G_C)=\frac{P(G_{\bar C}\cap G_C)}{P(G_{\bar C}\cap G_C)+P(G_{\bar C}\cap R_C)}=\frac{\frac29}{\frac49}=\frac12$
Just go with the flow. Even I was stumped a bit by the $2^{nd}$ part of the question, read it $3$ times and ... done!