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 of rolling a number greater than or equal to 4 on a loaded die, where each odd number is twice as likely to occur as each even number.

### Step-by-Step Solution:

1. **Define the Sample Space**:
The sample space for a die consists of the numbers {1, 2, 3, 4, 5, 6}.

2. **Assign Probabilities**:
Let the probability of rolling an even number (2, 4, 6) be \( p \). Since each odd number (1, 3, 5) is twice as likely to occur, the probability of rolling an odd number will be \( 2p \).

- For even numbers:
- P(2) = p
- P(4) = p
- P(6) = p
- For odd numbers:
- P(1) = 2p
- P(3) = 2p
- P(5) = 2p

3. **Set Up the Equation**:
The total probability must equal 1:
\[ P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 \]
Substituting the probabilities:
\[ 2p + p + 2p + p + 2p + p = 1 \]
Simplifying this gives:
\[ 9p = 1 \implies p = \frac{1}{9} \]

4. **Calculate Individual Probabilities**:
Now we can find the probabilities of each outcome:
- 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 Event E**:
The event E is defined as rolling a number greater than or equal to 4. The favorable outcomes for E are {4, 5, 6}.

6. **Calculate the Probability of Event E**:
The probability of event E is given by:
\[ P(E) = P(4) + P(5) + P(6) \]
Substituting the probabilities:
\[ P(E) = \frac{1}{9} + \frac{2}{9} + \frac{1}{9} = \frac{4}{9} \]

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