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)

We are tasked with finding the probability that the second urn was selected, given that a white ball was drawn. This is a classic problem of conditional probability, and we can solve it using **Bayes' Theorem**.

### Step 1: Define the events
Let:
- \( A_1 \) be the event that the first urn is selected.
- \( A_2 \) be the event that the second urn is selected.
- \( B \) be the event that a white ball is drawn.

We are given:
- \( P(A_1) = 0.65 \) (probability of selecting the first urn),
- \( P(A_2) = 1 - P(A_1) = 0.35 \) (probability of selecting the second urn),
- From the first urn, the probability of drawing a white ball \( P(B|A_1) \) is the ratio of white balls to total balls in the first urn: 
  \[
  P(B|A_1) = \frac{8}{8+2} = \frac{8}{10} = 0.8
  \]
- From the second urn, the probability of drawing a white ball \( P(B|A_2) \) is the ratio of white balls to total balls in the second urn:
  \[
  P(B|A_2) = \frac{3}{3+8} = \frac{3}{11} \approx 0.2727
  \]

### Step 2: Apply Bayes' Theorem
We are asked to find \( P(A_2 | B) \), the probability that the second urn was selected given that a white ball was drawn. Using Bayes' Theorem:
\[
P(A_2 | B) = \frac{P(B | A_2) P(A_2)}{P(B)}
\]
Where \( P(B) \), the total probability of drawing a white ball, is calculated as:
\[
P(B) = P(B | A_1) P(A_1) + P(B | A_2) P(A_2)
\]
Substitute the known values:
\[
P(B) = (0.8 \times 0.65) + (0.2727 \times 0.35) = 0.52 + 0.0954 = 0.6154
\]

### Step 3: Calculate \( P(A_2 | B) \)
Now, we calculate \( P(A_2 | B) \):
\[
P(A_2 | B) = \frac{0.2727 \times 0.35}{0.6154} = \frac{0.0954}{0.6154} \approx 0.1550
\]

Thus, the probability that the second urn was selected given that a white ball was drawn is approximately \( \boxed{0.1550} \).

Would you like any more details or further clarification?

Here are 5 related questions:
1. What is Bayes' Theorem, and when is it applied in probability?
2. How would the answer change if the probability of selecting the first urn was 0.5?
3. What is the significance of conditional probability in real-world applications?
4. How do we interpret \( P(B|A_1) \) and \( P(B|A_2) \) in this context?
5. Can this problem be solved without Bayes' Theorem? How would the approach differ?

**Tip:** Always break down complex probability problems into smaller events and steps, making sure to carefully apply relevant theorems such as Bayes' theorem or the law of total probability.

Solve the urn problem using Bayes' Theorem, where an urn is selected at random, and a ball is drawn. Given the probability of selecting each urn, learn how to calculate the probability that the second urn was chosen when a white ball is drawn. Suitable for students in Grades 11-12 or college.

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