Question 1166353
This is a classic problem that can be solved using **Bayes' Theorem**. We'll calculate the probability in two steps: after the first draw, and after the second draw.

## 🪨 Bag Contents and Initial Probabilities

First, let's define the contents of the two possible bags, $B_1$ and $B_2$, and the initial probabilities.

| Bag | Shiny Rocks ($S$) | Rough Rocks ($R$) | Total Rocks ($N$) |
| :---: | :---: | :---: | :---: |
| **B1** | 15 | 35 | 50 |
| **B2** | 40 | 10 | 50 |

Since you forgot which bag is which, the **prior probability** of selecting either bag is equal:
$$P(B_1) = P(B_2) = \frac{1}{2}$$

The probability of drawing a **Rough ($R$)** stone from each bag is:
* $P(R | B_1) = \frac{\text{Rough in B1}}{\text{Total in B1}} = \frac{35}{50} = \mathbf{0.7}$
* $P(R | B_2) = \frac{\text{Rough in B2}}{\text{Total in B2}} = \frac{10}{50} = \mathbf{0.2}$

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## 1. After the First Draw (A Rough Stone is Drawn)

Let $D_1$ be the event of drawing a **Rough** stone on the first draw. We want to find the **posterior probability** $P(B_1 | D_1)$.

### Step A: Probability of the Observation $P(D_1)$

The probability of drawing a rough stone from the *selected bag* is given by the Law of Total Probability:
$$P(D_1) = P(D_1 | B_1) \cdot P(B_1) + P(D_1 | B_2) \cdot P(B_2)$$
$$P(D_1) = (0.7) \cdot (\frac{1}{2}) + (0.2) \cdot (\frac{1}{2}) = 0.35 + 0.10 = \mathbf{0.45}$$

### Step B: Applying Bayes' Theorem

Bayes' Theorem states:
$$P(B_1 | D_1) = \frac{P(D_1 | B_1) \cdot P(B_1)}{P(D_1)}$$
$$P(B_1 | D_1) = \frac{(0.7) \cdot (0.5)}{0.45} = \frac{0.35}{0.45} = \frac{35}{45} = \frac{7}{9}$$

The probability that the bag you selected was bag $B_1$ after the first rough stone is $\mathbf{\frac{7}{9}}$ or approximately $\mathbf{0.7778}$.

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## 2. After the Second Draw (A Second Rough Stone is Drawn)

The first stone was *not replaced*. This means the total number of stones and the number of rough stones in the selected bag have both decreased by one.

Let $D_2$ be the event of drawing a **Rough** stone on the second draw *from the same bag*, given the first was rough. We want to find $P(B_1 | D_1 \text{ and } D_2)$.

### Step A: New Conditional Probabilities

The prior probabilities for the second step are the posterior probabilities from the first step:
* $P(B_1) = \frac{7}{9}$
* $P(B_2) = 1 - \frac{7}{9} = \frac{2}{9}$

Now we need the probability of drawing a second rough stone, $P(D_2)$, **given that a rough stone was already removed** (non-replacement).

| Bag | Initial $R$ | Initial $N$ | New $R$ | New $N$ | $P(D_2 | B_i)$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| **B1** | 35 | 50 | 34 | 49 | $\mathbf{\frac{34}{49}}$ (approx. 0.6939) |
| **B2** | 10 | 50 | 9 | 49 | $\mathbf{\frac{9}{49}}$ (approx. 0.1837) |

### Step B: Probability of the Second Observation $P(D_2)$

Using the Law of Total Probability with the new prior probabilities:
$$P(D_2) = P(D_2 | B_1) \cdot P(B_1) + P(D_2 | B_2) \cdot P(B_2)$$
$$P(D_2) = \left(\frac{34}{49}\right) \cdot \left(\frac{7}{9}\right) + \left(\frac{9}{49}\right) \cdot \left(\frac{2}{9}\right)$$
$$P(D_2) = \frac{34 \cdot 7}{49 \cdot 9} + \frac{9 \cdot 2}{49 \cdot 9} = \frac{238}{441} + \frac{18}{441} = \frac{256}{441}$$

### Step C: Applying Bayes' Theorem Again

The new posterior probability $P(B_1 | D_1 \text{ and } D_2)$ is:
$$P(B_1 | D_2) = \frac{P(D_2 | B_1) \cdot P(B_1)}{P(D_2)}$$
$$P(B_1 | D_2) = \frac{\left(\frac{34}{49}\right) \cdot \left(\frac{7}{9}\right)}{\frac{256}{441}} = \frac{\frac{238}{441}}{\frac{256}{441}}$$
$$P(B_1 | D_2) = \frac{238}{256} = \frac{119}{128}$$

The probability that the bag you selected is $B_1$ after the second rough stone (without replacement) is $\mathbf{\frac{119}{128}}$ or approximately $\mathbf{0.9297}$.

Would you like to see how the probability would have changed if the second stone had been a shiny stone instead?