Math Problem Statement

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)

Solution

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:

  • A1A_1 be the event that the first urn is selected.
  • A2A_2 be the event that the second urn is selected.
  • BB be the event that a white ball is drawn.

We are given:

  • P(A1)=0.65P(A_1) = 0.65 (probability of selecting the first urn),
  • P(A2)=1P(A1)=0.35P(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(BA1)P(B|A_1) is the ratio of white balls to total balls in the first urn: P(BA1)=88+2=810=0.8P(B|A_1) = \frac{8}{8+2} = \frac{8}{10} = 0.8
  • From the second urn, the probability of drawing a white ball P(BA2)P(B|A_2) is the ratio of white balls to total balls in the second urn: P(BA2)=33+8=3110.2727P(B|A_2) = \frac{3}{3+8} = \frac{3}{11} \approx 0.2727

Step 2: Apply Bayes' Theorem

We are asked to find P(A2B)P(A_2 | B), the probability that the second urn was selected given that a white ball was drawn. Using Bayes' Theorem: P(A2B)=P(BA2)P(A2)P(B)P(A_2 | B) = \frac{P(B | A_2) P(A_2)}{P(B)} Where P(B)P(B), the total probability of drawing a white ball, is calculated as: P(B)=P(BA1)P(A1)+P(BA2)P(A2)P(B) = P(B | A_1) P(A_1) + P(B | A_2) P(A_2) Substitute the known values: P(B)=(0.8×0.65)+(0.2727×0.35)=0.52+0.0954=0.6154P(B) = (0.8 \times 0.65) + (0.2727 \times 0.35) = 0.52 + 0.0954 = 0.6154

Step 3: Calculate P(A2B)P(A_2 | B)

Now, we calculate P(A2B)P(A_2 | B): P(A2B)=0.2727×0.350.6154=0.09540.61540.1550P(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 0.1550\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(BA1)P(B|A_1) and P(BA2)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.

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Math Problem Analysis

Mathematical Concepts

Probability
Conditional Probability
Bayes' Theorem

Formulas

P(A | B) = (P(B | A) * P(A)) / P(B)
P(B) = P(B | A_1) * P(A_1) + P(B | A_2) * P(A_2)

Theorems

Bayes' Theorem
Law of Total Probability

Suitable Grade Level

Grades 11-12 or College