A die is loaded so that the probability of a face I is proportional to I, I = 1, 2,………,6. Then the probability of an even number occurring when the die is rolled.

Options

A. `(2)/(7)`
B. `(1)/(7)`
C. `(3)/(7)`
D. `(4)/(7)`

Answer

To solve the problem, we need to determine the probability of rolling an even number on a loaded die, where the probability of each face \( i \) (where \( i = 1, 2, \ldots, 6 \)) is proportional to \( i \).

### Step-by-Step Solution:

1. **Define the Probabilities**:
Since the probability of face \( i \) is proportional to \( i \), we can express the probabilities as follows:
- Let the probability of rolling a 1 be \( P(1) = x \).
- Then, the probabilities for the other faces will be:
- \( P(2) = 2x \)
- \( P(3) = 3x \)
- \( P(4) = 4x \)
- \( P(5) = 5x \)
- \( P(6) = 6x \)

2. **Set Up the Total Probability**:
The total probability must equal 1:
\[ P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 \]
Substituting the expressions for the probabilities:
\[ x + 2x + 3x + 4x + 5x + 6x = 1 \]
Simplifying this gives:
\[ 21x = 1 \]

3. **Solve for \( x \)**:
To find \( x \), we divide both sides by 21:
\[ x = \frac{1}{21} \]

4. **Calculate the Probabilities of Even Numbers**:
The even numbers on the die are 2, 4, and 6. We can now find their probabilities:
- \( P(2) = 2x = 2 \times \frac{1}{21} = \frac{2}{21} \)
- \( P(4) = 4x = 4 \times \frac{1}{21} = \frac{4}{21} \)
- \( P(6) = 6x = 6 \times \frac{1}{21} = \frac{6}{21} \)

5. **Sum the Probabilities of Even Numbers**:
Now, we add the probabilities of the even numbers:
\[ P(\text{even}) = P(2) + P(4) + P(6) = \frac{2}{21} + \frac{4}{21} + \frac{6}{21} \]
Combining these fractions:
\[ P(\text{even}) = \frac{2 + 4 + 6}{21} = \frac{12}{21} \]

6. **Simplify the Result**:
Finally, we simplify \( \frac{12}{21} \):
\[ P(\text{even}) = \frac{4}{7} \]

### Final Answer:
The probability of rolling an even number when the die is rolled is \( \frac{4}{7} \).