An urn contains 5 white balls and 3 red balls. A second urn contains 4 white balls and 8 red balls. An urn is selected, and a ball is randomly drawn from the selected urn. The probability of selecting the first urn is 0.52. If the ball is white, find the probability that the first urn was selected. (Enter answer as a decimal with at least 4 correct decimal places)

To solve this problem, we need to apply **Bayes' Theorem**, which helps us find the probability of an event given the occurrence of another event.

We are asked to find the probability that the first urn was selected given that the ball drawn is white. Let's define the following events:

- \( A_1 \): The event that the first urn is selected.
- \( A_2 \): The event that the second urn is selected.
- \( B \): The event that a white ball is drawn.

### Step 1: Bayes' Theorem Formula

Bayes' Theorem states:
\[
P(A_1 | B) = \frac{P(B | A_1) \cdot P(A_1)}{P(B)}
\]
Where:
- \( P(A_1 | B) \) is the probability that the first urn was selected given that a white ball was drawn.
- \( P(B | A_1) \) is the probability of drawing a white ball given that the first urn was selected.
- \( P(A_1) \) is the prior probability of selecting the first urn.
- \( P(B) \) is the total probability of drawing a white ball.

### Step 2: Find the Individual Probabilities

#### 1. \( P(A_1) \): The probability that the first urn is selected is given as 0.52.
\[
P(A_1) = 0.52
\]

#### 2. \( P(A_2) \): The probability that the second urn is selected is the complement of \( P(A_1) \):
\[
P(A_2) = 1 - P(A_1) = 1 - 0.52 = 0.48
\]

#### 3. \( P(B | A_1) \): The probability of drawing a white ball given that the first urn is selected. The first urn contains 5 white balls and 3 red balls, so:
\[
P(B | A_1) = \frac{5}{5 + 3} = \frac{5}{8}
\]

#### 4. \( P(B | A_2) \): The probability of drawing a white ball given that the second urn is selected. The second urn contains 4 white balls and 8 red balls, so:
\[
P(B | A_2) = \frac{4}{4 + 8} = \frac{4}{12} = \frac{1}{3}
\]

### Step 3: Calculate \( P(B) \) (The Total Probability of Drawing a White Ball)

The total probability of drawing a white ball, \( P(B) \), is the weighted sum of the probabilities of drawing a white ball from either urn. Using the law of total probability:
\[
P(B) = P(B | A_1) \cdot P(A_1) + P(B | A_2) \cdot P(A_2)
\]
Substitute the known values:
\[
P(B) = \left(\frac{5}{8}\right) \cdot 0.52 + \left(\frac{1}{3}\right) \cdot 0.48
\]
\[
P(B) = \frac{5}{8} \cdot 0.52 + \frac{1}{3} \cdot 0.48 = 0.325 + 0.16 = 0.485
\]

### Step 4: Apply Bayes' Theorem

Now we can apply Bayes' Theorem to find \( P(A_1 | B) \):
\[
P(A_1 | B) = \frac{P(B | A_1) \cdot P(A_1)}{P(B)} = \frac{\left(\frac{5}{8}\right) \cdot 0.52}{0.485}
\]
\[
P(A_1 | B) = \frac{0.325}{0.485} \approx 0.6701
\]

Thus, the probability that the first urn was selected given that the ball drawn is white is approximately **0.6701**.

### Do you have any further questions, or would you like a detailed breakdown of any step?

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Here are 5 related questions to expand your understanding:

1. What would be the probability of selecting the second urn if a red ball was drawn?
2. How would the solution change if both urns had the same number of white and red balls?
3. What is the probability of drawing a red ball from the first urn?
4. How does the prior probability of selecting an urn affect the outcome in Bayes' Theorem?
5. How can you apply Bayes' Theorem to real-world problems outside of probability?

**Tip:** Bayes' Theorem is particularly useful when updating probabilities based on new information. It's commonly used in fields like medicine, finance, and artificial intelligence!

Solve a probability problem using Bayes' Theorem, involving two urns with different combinations of white and red balls. Learn how to calculate the probability that the first urn was selected given that a white ball was drawn. Includes a step-by-step breakdown suitable for students in Grades 11-12 or early college.

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