Question 1166353: You are recently separated your rock collection into two bags. You organize the bags so that one (bag B1) contains 15 shiny rocks and 35 rough rocks, and the other (bag B2) contains 40 shiny rocks and 10 rough rocks. You forgot which bag contains which set of rocks, and you would like to solve this problem by looking at as few stones as possible. You removed a random stone out of a bag which was a rough stone. What us the probability that the bag you selected was bag B1?
You decided to remove a second one to strengthen your confidence on which set the bag contains. You leave the first stone out of the bag and drawn another stone at random form the same bag. It was a rough stone again. What us the probability now that the bag you have been selecting stones from is bag B1.
Answer by CPhill(2030)   (Show Source):
You can put this solution on YOUR website! 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:
* $P(R | B_1) = \frac{\text{Rough in B1}}{\text{Total in B1}} = \frac{35}{50} = \mathbf{0.7}$
* $P(R | B_2) = \frac{\text{Rough in B2}}{\text{Total in B2}} = \frac{10}{50} = \mathbf{0.2}$
---
## 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 $\mathbf{0.7778}$.
---
## 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?
|
|
|
