A bag A contains 2 white and 3 red balls another bag B contains 4 white and 5 red balls. A bag is chosen at random and a ball is drawn from it. If the ball drawn is red, what is the probability that the bag B is chosen? [CBSE '04C)

Options

A. `5/14`
B. `5/16`
C. `5/18`
D. `25/52`

Answer

To solve the problem, we will use Bayes' theorem. Let's break down the solution step by step.

### Step 1: Define the Events
Let:
- \( A \): Event of choosing Bag A
- \( B \): Event of choosing Bag B
- \( R \): Event of drawing a red ball

### Step 2: Determine the Probabilities of Choosing Each Bag
Since a bag is chosen at random, the probabilities of choosing each bag are:
\[ P(A) = P(B) = \frac{1}{2} \]

### Step 3: Calculate the Probability of Drawing a Red Ball from Each Bag
- For Bag A, which contains 2 white and 3 red balls (total 5 balls):
\[ P(R | A) = \frac{\text{Number of red balls in A}}{\text{Total number of balls in A}} = \frac{3}{5} \]

- For Bag B, which contains 4 white and 5 red balls (total 9 balls):
\[ P(R | B) = \frac{\text{Number of red balls in B}}{\text{Total number of balls in B}} = \frac{5}{9} \]

### Step 4: Calculate the Total Probability of Drawing a Red Ball
Using the law of total probability:
\[ P(R) = P(R | A) \cdot P(A) + P(R | B) \cdot P(B) \]
Substituting the values:
\[ P(R) = \left(\frac{3}{5} \cdot \frac{1}{2}\right) + \left(\frac{5}{9} \cdot \frac{1}{2}\right) \]
Calculating each term:
\[ P(R) = \frac{3}{10} + \frac{5}{18} \]
To add these fractions, we need a common denominator (which is 90):
\[ P(R) = \frac{27}{90} + \frac{25}{90} = \frac{52}{90} = \frac{26}{45} \]

### Step 5: Apply Bayes' Theorem
We want to find \( P(B | R) \), the probability that Bag B was chosen given that a red ball was drawn:
\[ P(B | R) = \frac{P(R | B) \cdot P(B)}{P(R)} \]
Substituting the values:
\[ P(B | R) = \frac{\left(\frac{5}{9}\right) \cdot \left(\frac{1}{2}\right)}{\frac{26}{45}} \]
Calculating the numerator:
\[ P(B | R) = \frac{\frac{5}{18}}{\frac{26}{45}} = \frac{5}{18} \cdot \frac{45}{26} = \frac{225}{468} \]
Simplifying:
\[ P(B | R) = \frac{25}{52} \]

### Final Answer
The probability that Bag B was chosen given that a red ball was drawn is:
\[ \boxed{\frac{25}{52}} \]