A six-faced dice is so biased that it is twice as likely to show an even number as an odd number when thrown. It is thrown twice, the probability that the sum of two numbers thrown is even is `1//12` b. `1//6` c. `1//3` d. `5//9`

Options

A. `1//12`
B. `1//6`
C. `1//3`
D. `5//9`

Explanation

To solve the problem, we need to determine the probability that the sum of two numbers thrown on a biased six-faced die is even.

### Step-by-Step Solution:

1. **Define the Probabilities**:
Let the probability of rolling an odd number be \( P \). Since the die is biased to show even numbers twice as likely as odd numbers, the probability of rolling an even number will be \( 2P \).

2. **Set Up the Equation**:
Since the total probability must equal 1, we can write the equation:
\[ P + 2P = 1 \]
Simplifying this gives:
\[ 3P = 1 \]
Therefore, we find:
\[ P = \frac{1}{3} \]

3. **Calculate the Probability of Even and Odd Rolls**:
Now we can calculate the probabilities:
- Probability of rolling an odd number \( P(\text{odd}) = P = \frac{1}{3} \)
- Probability of rolling an even number \( P(\text{even}) = 2P = 2 \times \frac{1}{3} = \frac{2}{3} \)

4. **Determine Conditions for Even Sum**:
The sum of two numbers is even if:
- Both numbers are even (even + even)
- Both numbers are odd (odd + odd)

5. **Calculate the Probability for Each Case**:
- **Case 1: Even + Even**:
\[ P(\text{even + even}) = P(\text{even}) \times P(\text{even}) = \frac{2}{3} \times \frac{2}{3} = \frac{4}{9} \]

- **Case 2: Odd + Odd**:
\[ P(\text{odd + odd}) = P(\text{odd}) \times P(\text{odd}) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} \]

6. **Combine the Probabilities**:
Now, we add the probabilities of both cases to find the total probability of getting an even sum:
\[ P(\text{even sum}) = P(\text{even + even}) + P(\text{odd + odd}) = \frac{4}{9} + \frac{1}{9} = \frac{5}{9} \]

### Final Answer:
The probability that the sum of the two numbers thrown is even is:
\[ \frac{5}{9} \]