A die is rolled and showing any number is directly proportional to that number. If prime number appears then a ball is chosen from urn `A` containing 2 white and 3 black balls otherwise a ball is chosen from urn `B` containing 3 white and 2 black balls Then.
If white ball is draw then the probability that it is from urn `B`

Options

A. `52/53`
B. `1/53`
C. `20/53`
D. `33/53`

Explanation

To solve the problem step by step, we will use the concepts of probability and Bayes' theorem.

### Step 1: Determine the probabilities of rolling a die
The probability of rolling a number on a die is directly proportional to that number. Therefore, we can express the probabilities as follows:

- Let \( P(1) = k \)
- Let \( P(2) = 2k \)
- Let \( P(3) = 3k \)
- Let \( P(4) = 4k \)
- Let \( P(5) = 5k \)
- Let \( P(6) = 6k \)

The sum of all probabilities must equal 1:

\[ k + 2k + 3k + 4k + 5k + 6k = 1 \]

### Step 2: Solve for \( k \)
Combining the terms gives:

\[ 21k = 1 \implies k = \frac{1}{21} \]

Now we can find the individual probabilities:

- \( P(1) = \frac{1}{21} \)
- \( P(2) = \frac{2}{21} \)
- \( P(3) = \frac{3}{21} \)
- \( P(4) = \frac{4}{21} \)
- \( P(5) = \frac{5}{21} \)
- \( P(6) = \frac{6}{21} \)

### Step 3: Identify the prime numbers
The prime numbers on a die are 2, 3, and 5. We need to calculate the probability of rolling a prime number:

\[ P(\text{prime}) = P(2) + P(3) + P(5) = \frac{2}{21} + \frac{3}{21} + \frac{5}{21} = \frac{10}{21} \]

### Step 4: Calculate the probability of rolling a non-prime number
The probability of rolling a non-prime number is:

\[ P(\text{not prime}) = 1 - P(\text{prime}) = 1 - \frac{10}{21} = \frac{11}{21} \]

### Step 5: Determine the outcomes based on the urns
- If a prime number is rolled, a ball is chosen from urn A (which contains 2 white and 3 black balls).
- If a non-prime number is rolled, a ball is chosen from urn B (which contains 3 white and 2 black balls).

### Step 6: Calculate the probability of drawing a white ball from each urn
- From urn A, the probability of drawing a white ball is:

\[ P(W | A) = \frac{2}{5} \]

- From urn B, the probability of drawing a white ball is:

\[ P(W | B) = \frac{3}{5} \]

### Step 7: Use Bayes' theorem to find \( P(B | W) \)
We want to find the probability that the white ball drawn came from urn B, given that a white ball was drawn:

\[ P(B | W) = \frac{P(W | B) \cdot P(B)}{P(W)} \]

Where:
- \( P(B) = P(\text{not prime}) = \frac{11}{21} \)
- \( P(W) = P(W | A) \cdot P(A) + P(W | B) \cdot P(B) \)

Calculating \( P(W) \):

\[ P(W) = P(W | A) \cdot P(\text{prime}) + P(W | B) \cdot P(\text{not prime}) \]

Substituting the values:

\[ P(W) = \left(\frac{2}{5} \cdot \frac{10}{21}\right) + \left(\frac{3}{5} \cdot \frac{11}{21}\right) \]

Calculating each term:

\[ = \frac{20}{105} + \frac{33}{105} = \frac{53}{105} \]

### Step 8: Substitute back to find \( P(B | W) \)
Now substituting back into Bayes' theorem:

\[ P(B | W) = \frac{\left(\frac{3}{5}\right) \cdot \left(\frac{11}{21}\right)}{\frac{53}{105}} \]

Calculating the numerator:

\[ = \frac{33}{105} \]

So,

\[ P(B | W) = \frac{33/105}{53/105} = \frac{33}{53} \]

### Final Answer
The probability that the white ball drawn is from urn B is:

\[ \boxed{\frac{33}{53}} \]