A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 (ii) 12.

Explanation

To solve the problem, we first need to understand the possible outcomes when tossing a fair coin and a fair die.

### Step 1: Identify the outcomes

1. **Coin Outcomes**: The coin has two faces: one marked with 1 and the other marked with 6. Therefore, the possible outcomes from tossing the coin are:
- 1 (when the coin shows the face marked with 1)
- 6 (when the coin shows the face marked with 6)

2. **Die Outcomes**: A fair die has six faces numbered from 1 to 6. Therefore, the possible outcomes from tossing the die are:
- 1, 2, 3, 4, 5, 6

### Step 2: Calculate total outcomes

The total number of outcomes when tossing both the coin and the die can be calculated by multiplying the number of outcomes for the coin and the die:
- Total outcomes = Number of coin outcomes × Number of die outcomes = 2 × 6 = 12.

### Step 3: Find the favorable outcomes for each case

#### (i) Probability that the sum is 3

To find the combinations that yield a sum of 3, we can analyze the possible outcomes:

- If the coin shows **1**:
- The die must show **2** (1 + 2 = 3).

- If the coin shows **6**:
- The die cannot show any number that makes the sum equal to 3 (since 6 + any die number will exceed 3).

Thus, the only favorable outcome for a sum of 3 is:
- (Coin: 1, Die: 2)

**Number of favorable outcomes for sum = 3**: 1

#### (ii) Probability that the sum is 12

To find the combinations that yield a sum of 12, we analyze the possible outcomes:

- If the coin shows **1**:
- The die must show **11** (which is impossible since the die only goes up to 6).

- If the coin shows **6**:
- The die must show **6** (6 + 6 = 12).

Thus, the only favorable outcome for a sum of 12 is:
- (Coin: 6, Die: 6)

**Number of favorable outcomes for sum = 12**: 1

### Step 4: Calculate probabilities

Now we can calculate the probabilities for each case.

1. **Probability of sum = 3**:
\[ P(\text{sum} = 3) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{12} \]

2. **Probability of sum = 12**:
\[ P(\text{sum} = 12) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{12} \]

### Final Answers:
- (i) The probability that the sum is 3 is \( \frac{1}{12} \).
- (ii) The probability that the sum is 12 is \( \frac{1}{12} \).