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 find the probability of the event E, which is defined as rolling a number greater than or equal to 4 on a loaded die. The loading of the die means that odd numbers are more likely to occur than even numbers.

### Step-by-Step Solution:

1. **Define the Sample Space**:
The sample space S for a standard die consists of the numbers {1, 2, 3, 4, 5, 6}. However, since the die is loaded, we need to adjust the probabilities for each number.

2. **Assign Probabilities**:
Let's denote the probability of rolling an even number (2, 4, 6) as p. Since each odd number (1, 3, 5) is twice as likely to occur, the probabilities for the odd numbers will be 2p.
- Probability of rolling 1 = 2p
- Probability of rolling 2 = p
- Probability of rolling 3 = 2p
- Probability of rolling 4 = p
- Probability of rolling 5 = 2p
- Probability of rolling 6 = p

3. **Set Up the Total Probability**:
The total probability must sum to 1:
\[ 2p + p + 2p + p + 2p + p = 1 \]
Simplifying this gives:
\[ 9p = 1 \implies p = \frac{1}{9} \]

4. **Calculate Individual Probabilities**:
Now we can find the probabilities for each number:
- P(1) = 2p = 2 * (1/9) = 2/9
- P(2) = p = 1/9
- P(3) = 2p = 2/9
- P(4) = p = 1/9
- P(5) = 2p = 2/9
- P(6) = p = 1/9

5. **Identify Favorable Outcomes for Event E**:
The event E is rolling a number greater than or equal to 4. The favorable outcomes are {4, 5, 6}.
- P(4) = 1/9
- P(5) = 2/9
- P(6) = 1/9

6. **Calculate the Probability of Event E**:
To find P(E), we sum the probabilities of the favorable outcomes:
\[ P(E) = P(4) + P(5) + P(6) = \frac{1}{9} + \frac{2}{9} + \frac{1}{9} = \frac{4}{9} \]

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