An urn contains 8 white balls and 2 red balls. A second urn contains 3 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 0.65. If the ball is white, find the probability that the second urn was selected. (Enter answer as a decimal with at least 4 correct decimal places) make sure to round to the correct decimal place

This is a classic problem involving conditional probability and Bayes' Theorem.

We need to find the probability that the second 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.

### Given Information:
- \( P(A_1) = 0.65 \) (probability of selecting the first urn)
- \( P(A_2) = 1 - P(A_1) = 1 - 0.65 = 0.35 \) (probability of selecting the second urn)
- Probability of drawing a white ball from the first urn (\( P(B|A_1) \)):
  - The first urn contains 8 white balls out of a total of 10 balls (8 white + 2 red).
  - So, \( P(B|A_1) = \frac{8}{10} = 0.8 \).
  
- Probability of drawing a white ball from the second urn (\( P(B|A_2) \)):
  - The second urn contains 3 white balls out of a total of 11 balls (3 white + 8 red).
  - So, \( P(B|A_2) = \frac{3}{11} \approx 0.2727 \).

We want to find \( P(A_2 | B) \), the probability that the second urn was selected given that a white ball was drawn.

### Using Bayes' Theorem:

\[
P(A_2 | B) = \frac{P(A_2) P(B | A_2)}{P(A_1) P(B | A_1) + P(A_2) P(B | A_2)}
\]

Substitute the known values:

\[
P(A_2 | B) = \frac{0.35 \times 0.2727}{0.65 \times 0.8 + 0.35 \times 0.2727}
\]

Now, let's calculate this step-by-step.

### Step 1: Numerator
\[
0.35 \times 0.2727 \approx 0.095445
\]

### Step 2: Denominator
\[
0.65 \times 0.8 = 0.52
\]
\[
0.35 \times 0.2727 \approx 0.095445
\]
\[
0.52 + 0.095445 = 0.615445
\]

### Step 3: Final Calculation
\[
P(A_2 | B) = \frac{0.095445}{0.615445} \approx 0.1551
\]

Thus, the probability that the second urn was selected given that the ball is white is approximately **0.1551**.

### Do you want details or have any questions? Here are 5 related questions:
1. What is the general formula for Bayes' Theorem?
2. How would the answer change if the first urn contained more red balls?
3. What is the probability of drawing a red ball?
4. Can you derive the probability of selecting the first urn given that a red ball was drawn?
5. How does changing the probability of selecting the urns affect the overall outcome?

**Tip:** When applying Bayes' Theorem, make sure to carefully define the conditional probabilities and events for a clear understanding.

An urn contains 8 white balls and 2 red balls. A second urn contains 3 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 0.65. If the ball is white, find the probability that the second urn was selected. (Enter answer as a decimal with at least 4 correct decimal places)

Learn how to calculate the probability of selecting the second urn given that a white ball is drawn. This problem uses Bayes' Theorem and explores key concepts in conditional probability. The problem is ideal for students in Grades 11-12 or introductory college courses.

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