A die is loaded in such a way that each odd number is twice all likely to occur as each even number. If E is the event of a number greater than or equal to 4 on a single toss of the die, then P(E) is :

Options

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

Answer

To solve the problem, we need to determine the probability \( P(E) \) where event \( E \) is defined as rolling a number greater than or equal to 4 on a loaded die. The die is loaded such that each odd number is twice as likely to occur as each even number.

### Step-by-Step Solution:

1. **Identify the outcomes of a die roll:**
A standard die has six faces with numbers: 1, 2, 3, 4, 5, 6.

2. **Define the probabilities for odd and even numbers:**
Let the probability of rolling an even number (2, 4, 6) be \( p \). Since each odd number (1, 3, 5) is twice as likely as each even number, the probability for each odd number will be \( 2p \).

3. **Set up the total probability:**
The total probability must sum to 1:
\[ P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 \]
This can be expressed as:
\[ 2p + p + 2p + p + 2p + p = 1 \]
Simplifying this gives:
\[ 9p = 1 \quad \Rightarrow \quad p = \frac{1}{9} \]

4. **Calculate the probabilities for each number:**
- \( P(1) = 2p = \frac{2}{9} \)
- \( P(2) = p = \frac{1}{9} \)
- \( P(3) = 2p = \frac{2}{9} \)
- \( P(4) = p = \frac{1}{9} \)
- \( P(5) = 2p = \frac{2}{9} \)
- \( P(6) = p = \frac{1}{9} \)

5. **Identify the favorable outcomes for event \( E \):**
Event \( E \) includes the outcomes: 4, 5, and 6. Therefore, we need to find \( P(4) + P(5) + P(6) \):
\[ P(E) = P(4) + P(5) + P(6) = \frac{1}{9} + \frac{2}{9} + \frac{1}{9} = \frac{4}{9} \]

6. **Conclusion:**
The probability \( P(E) \) of rolling a number greater than or equal to 4 is:
\[ P(E) = \frac{4}{9} \]

### Final Answer:
The probability \( P(E) \) is \( \frac{4}{9} \).

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