Let a die be loaded in such a way that even faces are twice likely to occur as the odd faces. What is the probability that a prime number will show up when the die is tossed?

Options

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

Answer

To solve the problem, we need to determine the probability of rolling a prime number on a loaded die where even faces are twice as likely to occur as odd faces.

### Step-by-Step Solution:

1. **Identify the Faces of the Die**:
The faces of a standard die are 1, 2, 3, 4, 5, and 6. Among these, the prime numbers are 2, 3, and 5.

2. **Define Probabilities for Odd and Even Faces**:
Let the probability of rolling an odd face (1, 3, 5) be \( p \). Since even faces (2, 4, 6) are twice as likely, the probability of rolling an even face will be \( 2p \).

3. **Set Up the Total Probability Equation**:
The total probability must equal 1. There are 3 odd faces and 3 even faces, so we can express this as:
\[ 3p + 3(2p) = 1 \]
Simplifying this gives:
\[ 3p + 6p = 1 \implies 9p = 1 \implies p = \frac{1}{9} \]

4. **Calculate the Probabilities for Each Face**:
- Probability of rolling an odd face (1, 3, 5):
\[ P(1) = P(3) = P(5) = p = \frac{1}{9} \]
- Probability of rolling an even face (2, 4, 6):
\[ P(2) = P(4) = P(6) = 2p = 2 \times \frac{1}{9} = \frac{2}{9} \]

5. **Determine the Probability of Rolling a Prime Number**:
The prime numbers on the die are 2, 3, and 5. Thus, the probability of rolling a prime number is:
\[ P(\text{prime}) = P(2) + P(3) + P(5) = \frac{2}{9} + \frac{1}{9} + \frac{1}{9} \]
Simplifying this gives:
\[ P(\text{prime}) = \frac{2 + 1 + 1}{9} = \frac{4}{9} \]

6. **Final Answer**:
The probability that a prime number will show up when the die is tossed is \( \frac{4}{9} \).

### Summary of Steps:
1. Identify the faces of the die and the prime numbers.
2. Define the probabilities for odd and even faces.
3. Set up and solve the total probability equation.
4. Calculate the probabilities for each face.
5. Determine the probability of rolling a prime number.
6. State the final answer.