The probability of getting a prime number when a die is thrown once is `(2)/(3)`.

Answer

To determine if the statement "The probability of getting a prime number when a die is thrown once is \( \frac{2}{3} \)" is true or false, we can follow these steps:

### Step 1: Identify the total outcomes when a die is thrown.
When a standard die is thrown, the possible outcomes are the numbers 1, 2, 3, 4, 5, and 6. Therefore, the total number of outcomes is:

\[ \text{Total outcomes} = 6 \]

### Step 2: Identify the prime numbers among the outcomes.
The prime numbers between 1 and 6 are:
- 2 (only divisible by 1 and itself)
- 3 (only divisible by 1 and itself)
- 5 (only divisible by 1 and itself)

Thus, the prime numbers are 2, 3, and 5.

### Step 3: Count the favorable outcomes.
The favorable outcomes for getting a prime number are 2, 3, and 5. Therefore, the number of favorable outcomes is:

\[ \text{Favorable outcomes} = 3 \]

### Step 4: Calculate the probability of getting a prime number.
The probability \( P \) of an event is given by the formula:

\[ P(\text{event}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} \]

Substituting the values we found:

\[ P(\text{prime number}) = \frac{3}{6} = \frac{1}{2} \]

### Step 5: Compare the calculated probability with the given statement.
The statement claims that the probability of getting a prime number is \( \frac{2}{3} \). However, we calculated that the probability is \( \frac{1}{2} \).

### Conclusion:
Since \( \frac{1}{2} \) is not equal to \( \frac{2}{3} \), the given statement is false.

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