A coin is tossed 3 times. List the possible outcomes, find the probability of getting :
(i) all heads
(ii) at least two heads

Answer

To solve the problem, we will follow these steps:

### Step 1: List the Possible Outcomes
When a coin is tossed 3 times, the possible outcomes can be represented as combinations of Heads (H) and Tails (T). The total number of outcomes can be calculated as \(2^n\), where \(n\) is the number of tosses. Here, \(n = 3\).

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

The possible outcomes are:
1. HHH
2. HHT
3. HTH
4. THH
5. HTT
6. THT
7. TTH
8. TTT

### Step 2: Find the Probability of Getting All Heads
To find the probability of getting all heads (HHH), we need to identify the number of favorable outcomes and the total number of outcomes.

- **Favorable outcomes for all heads**: 1 (which is HHH)
- **Total outcomes**: 8

The probability \(P\) of getting all heads is given by:

\[ P(\text{All Heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{8} \]

### Step 3: Find the Probability of Getting At Least Two Heads
Next, we need to find the probability of getting at least two heads. This includes the outcomes with exactly 2 heads and the outcome with 3 heads.

- **Favorable outcomes for at least two heads**:
- Outcomes with exactly 2 heads: HHT, HTH, THH (3 outcomes)
- Outcome with 3 heads: HHH (1 outcome)

Thus, the total number of favorable outcomes for at least two heads is:

\[ 3 \text{ (for 2 heads)} + 1 \text{ (for 3 heads)} = 4 \]

The probability \(P\) of getting at least two heads is given by:

\[ P(\text{At least 2 Heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{8} = \frac{1}{2} \]

### Final Answers
(i) The probability of getting all heads is \(\frac{1}{8}\).

(ii) The probability of getting at least two heads is \(\frac{1}{2}\).

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