Bag A contains 2 white and 3 red marbles and bag B contains 4 white and 5 red marbles. One marble is drawn at random from one of the bags and is found to be red. The probability that it was drawn from the bag B is

Options

A. `4/13`
B. `25/52`
C. `36/65`
D. `41/78`

Answer

To solve the problem, we need to find the probability that a red marble drawn is from bag B, given that a red marble has been drawn. We can use Bayes' theorem to find this probability.

### Step 1: Define the events
Let:
- \( A \): the event that a marble is drawn from bag A.
- \( B \): the event that a marble is drawn from bag B.
- \( R \): the event that a red marble is drawn.

### Step 2: Find the probabilities of drawing a red marble from each bag
- Bag A contains 2 white and 3 red marbles, so the total number of marbles in bag A is \( 2 + 3 = 5 \).
- The probability of drawing a red marble from bag A is:
\[ P(R|A) = \frac{3}{5} \]

- Bag B contains 4 white and 5 red marbles, so the total number of marbles in bag B is \( 4 + 5 = 9 \).
- The probability of drawing a red marble from bag B is:
\[ P(R|B) = \frac{5}{9} \]

### Step 3: Find the total probability of drawing a red marble
To find the total probability of drawing a red marble from either bag, we need to consider the probabilities of choosing each bag. Assuming that each bag is equally likely to be chosen:
- The probability of choosing bag A is \( P(A) = \frac{1}{2} \).
- The probability of choosing bag B is \( P(B) = \frac{1}{2} \).

Now, we can use the law of total probability:
\[ P(R) = P(R|A)P(A) + P(R|B)P(B) \]
Substituting the values we have:
\[ P(R) = \left(\frac{3}{5} \cdot \frac{1}{2}\right) + \left(\frac{5}{9} \cdot \frac{1}{2}\right) \]
\[ P(R) = \frac{3}{10} + \frac{5}{18} \]

### Step 4: Find a common denominator and calculate \( P(R) \)
The least common multiple of 10 and 18 is 90. We convert both fractions:
\[ \frac{3}{10} = \frac{27}{90}, \quad \frac{5}{18} = \frac{25}{90} \]
Now, adding these:
\[ P(R) = \frac{27}{90} + \frac{25}{90} = \frac{52}{90} = \frac{26}{45} \]

### Step 5: Apply Bayes' theorem to find \( P(B|R) \)
Now we can find the probability that the red marble was drawn from bag B using Bayes' theorem:
\[ P(B|R) = \frac{P(R|B)P(B)}{P(R)} \]
Substituting the known values:
\[ P(B|R) = \frac{\left(\frac{5}{9}\right) \left(\frac{1}{2}\right)}{\frac{26}{45}} \]
Calculating the numerator:
\[ P(B|R) = \frac{\frac{5}{18}}{\frac{26}{45}} = \frac{5}{18} \cdot \frac{45}{26} = \frac{5 \cdot 45}{18 \cdot 26} \]
Simplifying:
\[ = \frac{225}{468} = \frac{25}{52} \]

### Final Answer
Thus, the probability that the red marble was drawn from bag B is:
\[ \boxed{\frac{25}{52}} \]