Two dice are rolled together and the numbers x, y on them form complex number z = x + iy. The probability that `|z| le 3` is

Options

A. `1/3`
B. `2/3`
C. `1/12`
D. `5/36`

Answer

To solve the problem, we need to find the probability that the modulus of the complex number \( z = x + iy \) formed by the numbers on two rolled dice satisfies the condition \( |z| \leq 3 \).

### Step-by-Step Solution:

1. **Understand the Problem**:
We are rolling two dice, which gives us two numbers \( x \) and \( y \). The complex number formed is \( z = x + iy \). We need to find the probability that \( |z| \leq 3 \).

2. **Total Outcomes**:
Each die has 6 faces, so the total number of outcomes when rolling two dice is:
\[ \text{Total outcomes} = 6 \times 6 = 36 \]

3. **Condition on Modulus**:
The modulus of the complex number \( z \) is given by:
\[ |z| = \sqrt{x^2 + y^2} \]
We want to find the cases where:
\[ |z| \leq 3 \implies \sqrt{x^2 + y^2} \leq 3 \]
Squaring both sides, we get:
\[ x^2 + y^2 \leq 9 \]

4. **Finding Favorable Outcomes**:
We will now find pairs \( (x, y) \) such that \( x^2 + y^2 \leq 9 \) with \( x \) and \( y \) taking values from 1 to 6 (since these are the possible outcomes of rolling a die).

- **For \( x = 1 \)**:
- \( 1^2 + y^2 \leq 9 \) → \( y^2 \leq 8 \) → Possible values: \( y = 1, 2, 3 \) (3 cases)
- **For \( x = 2 \)**:
- \( 2^2 + y^2 \leq 9 \) → \( y^2 \leq 5 \) → Possible values: \( y = 1, 2 \) (2 cases)
- **For \( x = 3 \)**:
- \( 3^2 + y^2 \leq 9 \) → \( y^2 \leq 0 \) → Only possible value: \( y = 1 \) (1 case)
- **For \( x = 4, 5, 6 \)**:
- \( 4^2 = 16 \), \( 5^2 = 25 \), \( 6^2 = 36 \) → All these exceed 9, so no valid pairs.

5. **Counting Favorable Cases**:
Now we can list the valid pairs:
- From \( x = 1 \): \( (1, 1), (1, 2), (1, 3) \) → 3 cases
- From \( x = 2 \): \( (2, 1), (2, 2) \) → 2 cases
- From \( x = 3 \): \( (3, 1) \) → 1 case

Total favorable cases = \( 3 + 2 + 1 = 6 \).

6. **Calculating Probability**:
The probability \( P \) that \( |z| \leq 3 \) is given by the ratio of favorable outcomes to total outcomes:
\[ P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{6}{36} = \frac{1}{6} \]

### Final Answer:
The probability that \( |z| \leq 3 \) is \( \frac{1}{6} \).