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 block balls otherwise a ball is chosen from urn B containing 3 white and 2 block balls then.
Q. 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 calculate the probability that a white ball drawn is from urn B, given the conditions of the problem.

### Step 1: Determine the probabilities of rolling each number on the die.
Since the probability of rolling a number is directly proportional to that number, we can express the probabilities as follows:
- Let \( P(X) \) be the probability of rolling \( X \) on the die.
- \( P(1) = k \)
- \( P(2) = 2k \)
- \( P(3) = 3k \)
- \( P(4) = 4k \)
- \( P(5) = 5k \)
- \( P(6) = 6k \)

The total probability must equal 1:
\[ k + 2k + 3k + 4k + 5k + 6k = 1 \]
\[ 21k = 1 \implies k = \frac{1}{21} \]

### Step 2: Calculate the probabilities for each outcome.
Using \( k = \frac{1}{21} \):
- \( P(1) = \frac{1}{21} \)
- \( P(2) = \frac{2}{21} \)
- \( P(3) = \frac{3}{21} \)
- \( P(4) = \frac{4}{21} \)
- \( P(5) = \frac{5}{21} \)
- \( P(6) = \frac{6}{21} \)

### Step 3: Identify the prime numbers on the die.
The prime numbers from 1 to 6 are 2, 3, and 5.

### Step 4: 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 5: Calculate the probability of rolling a non-prime number.
\[ P(\text{Non-prime}) = 1 - P(\text{Prime}) = 1 - \frac{10}{21} = \frac{11}{21} \]

### Step 6: Determine the probabilities of drawing a white ball from each urn.
- **Urn A** (chosen if a prime number is rolled): Contains 2 white and 3 black balls.
\[ P(\text{White} | A) = \frac{2}{5} \]

- **Urn B** (chosen if a non-prime number is rolled): Contains 3 white and 2 black balls.
\[ P(\text{White} | B) = \frac{3}{5} \]

### Step 7: Use Bayes' theorem to find the probability that the white ball is from urn B.
We want to find \( P(B | \text{White}) \):
\[ P(B | \text{White}) = \frac{P(\text{White} | B) \cdot P(B)}{P(\text{White})} \]

### Step 8: Calculate \( P(\text{White}) \).
Using the law of total probability:
\[ P(\text{White}) = P(\text{White} | A) \cdot P(A) + P(\text{White} | B) \cdot P(B) \]
\[ = \left(\frac{2}{5} \cdot \frac{10}{21}\right) + \left(\frac{3}{5} \cdot \frac{11}{21}\right) \]
\[ = \frac{20}{105} + \frac{33}{105} = \frac{53}{105} \]

### Step 9: Substitute back into Bayes' theorem.
\[ P(B | \text{White}) = \frac{\left(\frac{3}{5}\right) \cdot \left(\frac{11}{21}\right)}{\frac{53}{105}} \]
\[ = \frac{\frac{33}{105}}{\frac{53}{105}} = \frac{33}{53} \]

### Final Answer:
The probability that the white ball drawn is from urn B is \( \frac{33}{53} \).

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