Question

Bag A contains 2 white and 3 red balls, and Bag B contains 4 white and 5 red balls. If one ball is drawn at random from one of the bags and is found to be red, then the probability that it was drawn from Bag B is:

• A. \( \frac{23}{54} \)
• B. \( \frac{25}{51} \)
• C. \( \frac{25}{52} \)
• D. \( \frac{27}{55} \)

Hint:

Use Bayes’ theorem to find conditional probabilities when given dependent events.

Answer 1

The Correct Option is C

To solve this problem, we will use Bayes' Theorem, which helps us find the probability of an event given prior knowledge. Here, we want to find the probability that the red ball was drawn from Bag B, given that a red ball was drawn. Step 1: Define the events
Let \( A \) be the event that the ball is drawn from Bag A.
Let \( B \) be the event that the ball is drawn from Bag B.
Let \( R \) be the event that the ball drawn is red. We are tasked with finding \( P(B | R) \), the probability that the ball was drawn from Bag B, given that it is red. Step 2: Compute the probabilities
1. Probability of selecting a bag:
Since the bag is chosen at random, the probability of selecting Bag A or Bag B is equal: \[ P(A) = \frac{1}{2}, \quad P(B) = \frac{1}{2}. \] 2. Probability of drawing a red ball from each bag:
Bag A contains 2 white and 3 red balls, so the probability of drawing a red ball from Bag A is: \[ P(R | A) = \frac{3}{5}. \] Bag B contains 4 white and 5 red balls, so the probability of drawing a red ball from Bag B is: \[ P(R | B) = \frac{5}{9}. \] 3. Total probability of drawing a red ball:
Using the Law of Total Probability, the total probability of drawing a red ball is: \[ P(R) = P(R | A) \cdot P(A) + P(R | B) \cdot P(B). \] Substituting the values: \[ P(R) = \left(\frac{3}{5}\right) \cdot \left(\frac{1}{2}\right) + \left(\frac{5}{9}\right) \cdot \left(\frac{1}{2}\right). \] Simplify: \[ P(R) = \frac{3}{10} + \frac{5}{18}. \] To add these fractions, find a common denominator (90): \[ P(R) = \frac{27}{90} + \frac{25}{90} = \frac{52}{90} = \frac{26}{45}. \] Step 3: Apply Bayes' Theorem Bayes' Theorem states: \[ P(B | R) = \frac{P(R | B) \cdot P(B)}{P(R)}. \] Substitute the known values: \[ P(B | R) = \frac{\left(\frac{5}{9}\right) \cdot \left(\frac{1}{2}\right)}{\frac{26}{45}}. \] Simplify the numerator: \[ P(B | R) = \frac{\frac{5}{18}}{\frac{26}{45}}. \] Divide the fractions by multiplying by the reciprocal: \[ P(B | R) = \frac{5}{18} \cdot \frac{45}{26}. \] Simplify: \[ P(B | R) = \frac{5 \cdot 45}{18 \cdot 26} = \frac{225}{468}. \] Reduce the fraction by dividing numerator and denominator by 9: \[ P(B | R) = \frac{25}{52}. \] Final Answer: The probability that the red ball was drawn from Bag B is: \[ \boxed{\frac{25}{52}} \]