Let a die is loaded in such a way that prime number faces are twice as likely to occur as a non prime number faces. The probability that an odd number will be show up when die is tossed is-

Options

A. `(1)/(3)`
B. `(2)/(3)`
C. `(4)/(9)`
D. `(5)/(9)`

Answer

To solve the problem, we need to determine the probability that an odd number will show up when a loaded die is tossed, where prime number faces are twice as likely to occur as non-prime number faces.

### Step-by-Step Solution:

1. **Identify the faces of the die**: A standard die has six faces numbered 1, 2, 3, 4, 5, and 6. Among these:
- Prime numbers: 2, 3, 5
- Non-prime numbers: 1, 4, 6
- Odd numbers: 1, 3, 5

2. **Define probabilities**: Let the probability of a non-prime number face showing up be \( X \). Since prime number faces are twice as likely to occur, the probability of a prime number face showing up will be \( 2X \).

3. **Calculate total probabilities**: There are three non-prime faces and three prime faces on the die. Therefore, the total probability can be expressed as:
\[ \text{Total probability} = 3X + 3(2X) = 3X + 6X = 9X \]

4. **Set the total probability equal to 1**: Since the total probability must equal 1, we have:
\[ 9X = 1 \]
Solving for \( X \):
\[ X = \frac{1}{9} \]

5. **Calculate individual probabilities**:
- Probability of a non-prime number (1, 4, 6): \( P(1) = P(4) = P(6) = X = \frac{1}{9} \)
- Probability of a prime number (2, 3, 5): \( P(2) = P(3) = P(5) = 2X = \frac{2}{9} \)

6. **Calculate the probability of odd numbers**: The odd numbers on the die are 1, 3, and 5. Therefore, the probability of rolling an odd number is:
\[ P(\text{odd}) = P(1) + P(3) + P(5) = \frac{1}{9} + \frac{2}{9} + \frac{2}{9} \]
Simplifying this:
\[ P(\text{odd}) = \frac{1 + 2 + 2}{9} = \frac{5}{9} \]

7. **Final Result**: The probability that an odd number will show up when the die is tossed is:
\[ \boxed{\frac{5}{9}} \]