Three coins are tossed simultaneously. Write the sample space and complete the following table:
Number of Heads: `" "0" "1" "2" "3`
Probability: `" "(i)" "(ii)" (iii)" "(iv)`

Answer

To solve the problem, we will first determine the sample space when three coins are tossed simultaneously. Then, we will calculate the probabilities for different numbers of heads.

### Step 1: Write the Sample Space
When three coins are tossed, each coin has two possible outcomes: Heads (H) or Tails (T). The total number of outcomes when tossing three coins is calculated as follows:

\[ \text{Total Outcomes} = 2^3 = 8 \]

The sample space (S) consists of all possible outcomes:
- HHH
- HHT
- HTH
- THH
- HTT
- THT
- TTH
- TTT

So, the sample space is:
\[ S = \{ \text{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} \} \]

### Step 2: Count the Number of Heads
Now, we will count the number of heads in each outcome and fill in the table.

- **0 Heads**: Only TTT (1 outcome)
- **1 Head**: HHT, HTH, THH (3 outcomes)
- **2 Heads**: HHT, HTH, THH (3 outcomes)
- **3 Heads**: HHH (1 outcome)

### Step 3: Calculate the Probabilities
The probability of an event is given by the formula:
\[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \]

Now we can fill in the probabilities for each case:

1. **Probability of 0 Heads**:
\[ P(0 \text{ Heads}) = \frac{1}{8} \]

2. **Probability of 1 Head**:
\[ P(1 \text{ Head}) = \frac{3}{8} \]

3. **Probability of 2 Heads**:
\[ P(2 \text{ Heads}) = \frac{3}{8} \]

4. **Probability of 3 Heads**:
\[ P(3 \text{ Heads}) = \frac{1}{8} \]

### Final Table
Now we can complete the table:

| Number of Heads | Probability |
|-----------------|------------------|
| 0 | \( \frac{1}{8} \) |
| 1 | \( \frac{3}{8} \) |
| 2 | \( \frac{3}{8} \) |
| 3 | \( \frac{1}{8} \) |