This problem is a classic example of conditional probability. Here's the setup:

Problem Recap:

• You have 3 boxes:
  1. Box 1: Contains 2 red marbles (RR).
  2. Box 2: Contains 2 green marbles (GG).
  3. Box 3: Contains 1 red and 1 green marble (RG).
• A marble is drawn, and it's red. We want to know the probability that the other marble in the box is also red.

Solution Approach:

We are interested in finding the probability that the other marble in the box is also red given that the marble drawn is red. This is a conditional probability problem, so we'll use Bayes' Theorem.

We are given that a red marble was drawn, and we want the probability that the box chosen was Box 1 (which contains two red marbles).

Let’s define the events:

• A₁: The box chosen was Box 1 (RR).
• A₂: The box chosen was Box 2 (GG).
• A₃: The box chosen was Box 3 (RG).
• B: A red marble was drawn.

We need to find $P(A₁ \mid B)$, the probability that the box was Box 1 given that a red marble was drawn.

Using Bayes' Theorem:

$P(A₁ \mid B) = \frac{P(B \mid A₁) P(A₁)}{P(B)}$

Where:

• $P(A₁)$ is the prior probability of choosing any box, which is $\frac{1}{3}$ because there are three boxes and the choice is random.
• $P(B \mid A₁)$ is the probability of drawing a red marble given Box 1 was chosen. Since Box 1 contains 2 red marbles, this is 1 (since you are certain to draw a red marble).
• $P(B)$ is the total probability of drawing a red marble from any of the boxes. This can be calculated by considering the likelihood of drawing a red marble from each box: $P(B) = P(B \mid A₁)P(A₁) + P(B \mid A₂)P(A₂) + P(B \mid A₃)P(A₃)$
  • From Box 1: $P(B \mid A₁) = 1$.
  • From Box 2: $P(B \mid A₂) = 0$ (because there are no red marbles in Box 2).
  • From Box 3: $P(B \mid A₃) = \frac{1}{2}$ (because one of the two marbles in Box 3 is red).
  • Each of the boxes has a prior probability of $\frac{1}{3}$.

Thus, $P(B) = (1)(\frac{1}{3}) + (0)(\frac{1}{3}) + \left(\frac{1}{2}\right)(\frac{1}{3}) = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$

Now, applying Bayes' Theorem: $P(A₁ \mid B) = \frac{(1)(\frac{1}{3})}{\frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}$

Conclusion:

The probability that the other marble in the box is also red (i.e., the box was Box 1) given that a red marble was drawn is $\frac{2}{3}$.

Would you like more details on this solution?

Here are 5 related questions you might find interesting:

1. How does Bayes' Theorem work in conditional probability problems?
2. What if the problem asked for the probability that the drawn marble is green instead of red?
3. How do you calculate the probability of drawing a red marble from multiple boxes with different distributions?
4. What are common mistakes when using Bayes' Theorem?
5. Can this problem be extended to more boxes or different types of marbles?

Tip: When solving problems like this, carefully define the events and conditions. Writing out Bayes’ formula and interpreting the problem step-by-step helps avoid confusion.