A dice is thrown twice. Find the probability of getting 3 at least one time.

Explanation

To find the probability of getting a 3 at least once when a die is thrown twice, we can follow these steps:

### Step 1: Determine the total number of outcomes
When a die is thrown once, there are 6 possible outcomes (1, 2, 3, 4, 5, 6). Since the die is thrown twice, the total number of possible outcomes is:
\[ 6 \times 6 = 36 \]

### Step 2: Define the event of interest
Let \( E \) be the event of getting a 3 at least once in two throws.

### Step 3: Calculate the number of favorable outcomes for event \( E \)
To find the number of outcomes where at least one die shows a 3, we can list the combinations:
- If the first die shows a 3, the second die can show any of the 6 outcomes:
- (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) → 6 outcomes
- If the second die shows a 3, the first die can show any of the 6 outcomes:
- (1,3), (2,3), (4,3), (5,3), (6,3) → 5 additional outcomes (note that (3,3) is already counted)

Thus, the total number of favorable outcomes is:
\[ 6 + 5 = 11 \]

### Step 4: Calculate the probability
The probability \( P(E) \) of event \( E \) is given by the formula:
\[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{11}{36} \]

### Final Answer
The probability of getting a 3 at least once when a die is thrown twice is:
\[ \frac{11}{36} \]

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