If two fair dices are thrown and digits on dices are a and b, then find the probability for which `omega^(ab) = 1`, (where `omega` is a cube root of unity).

Explanation

To solve the problem of finding the probability that \( \omega^{ab} = 1 \) when two fair dice are thrown, we can follow these steps:

### Step 1: Understand the condition \( \omega^{ab} = 1 \)
The cube roots of unity are \( 1, \omega, \text{ and } \omega^2 \), where \( \omega = e^{2\pi i / 3} \). The condition \( \omega^{ab} = 1 \) holds true if \( ab \) is a multiple of 3. This means we need to find the pairs \( (a, b) \) such that \( ab \equiv 0 \mod 3 \).

### Step 2: Total outcomes when throwing two dice
When two dice are thrown, each die has 6 faces. Therefore, the total number of outcomes is:
\[ 6 \times 6 = 36 \]

### Step 3: Identify the pairs where \( ab \) is a multiple of 3
To satisfy \( ab \equiv 0 \mod 3 \), at least one of \( a \) or \( b \) must be a multiple of 3. The multiples of 3 on a die are 3 and 6.

### Step 4: Count the cases where neither die shows a multiple of 3
The numbers on the dice that are not multiples of 3 are 1, 2, 4, and 5. Thus, there are 4 options for each die:
- Die 1: 1, 2, 4, 5 (4 options)
- Die 2: 1, 2, 4, 5 (4 options)

The total number of outcomes where neither die shows a multiple of 3 is:
\[ 4 \times 4 = 16 \]

### Step 5: Calculate the cases where at least one die shows a multiple of 3
To find the number of cases where at least one die shows a multiple of 3, we subtract the cases where neither die shows a multiple of 3 from the total outcomes:
\[ 36 - 16 = 20 \]

### Step 6: Calculate the probability
The probability that \( ab \) is a multiple of 3 is given by the ratio of favorable outcomes to total outcomes:
\[ P(ab \equiv 0 \mod 3) = \frac{20}{36} = \frac{5}{9} \]

### Final Answer
Thus, the probability that \( \omega^{ab} = 1 \) is:
\[ \frac{5}{9} \]