Suppose a bag exactly 5 marbles that are either red or green, and the probability of the bag containing 0, 1, 2,…,5 red marbles is uniform (e.g., each has probability 1/6). One person draws a marble from the bag, and it is green. If this green marble is not put back into the bag, what is the probability that the next marble will be red?
I think this is a routine Bayes rule question. If A is the event that the next marble is red and B is the event that the first marble was green, I have:
$$P(A | B) = \frac{P(B | A)P(A)}{P(B)}$$
However, I'm not quite sure this makes sense, since B|A means that the first marble being green somehow depends on the second marble being red, which is silly because the first draw can't be impacted by the second draw because that would violate temporality. Can someone please help me figure out how to solve this problem?
Best Answer
You have, for $N_R$ the number of red marbles in the bag: $\mathsf P(N_R=r) = \begin{cases}\frac 1 6 & : r\in\{0,1,2,3,4,5\} \\ 0 & : \text{elsewhere}\end{cases}$
For any given value of $N_R=r$, we have:
Then use the Law of Total Probability to obtain the probability of drawing a green: $$\begin{align} \mathsf P(G_1) & = \sum_{r=0}^5 \mathsf P(G_1\mid N_R=r)\mathsf P(N_R=r) \\ & = \ldots \end{align}$$
Similarly the probability of drawing a green, and then a red on the next turn is: $$\begin{align} \mathsf P(G_1 \cap R_2) & = \sum_{r=0}^5 \mathsf P(G_1\cap R_2\mid N_R=r)\mathsf P(N_R=r) \\ & = \ldots \end{align}$$
So the conditional probability is: $$\begin{align} \mathsf P(R_2\mid G_1) & =\frac{\mathsf P(G_1\cap R_2)}{\mathsf P(G_1)} \\ & = \ldots \end{align}$$
Can you complete?