Let's solve the problem step by step:

Basic Setup:

A coin has two possible outcomes: Head (H) or Tail (T). When a coin is tossed 3 times, the total number of outcomes is $2^3 = 8$, since each toss has two possibilities.

The possible outcomes for three tosses are: $\{ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT \}$ There are 8 equally likely outcomes.

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(i) Probability of getting exactly one head:

We need to count the outcomes where exactly one head appears.

The outcomes with exactly one head are: $\{ HTT, THT, TTH \}$ There are 3 such outcomes.

So, the probability $P(\text{exactly one head})$ is: $P(\text{exactly one head}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{3}{8}$

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(ii) Probability of getting at least two heads:

The outcomes with at least two heads mean either 2 heads or 3 heads.

Outcomes with exactly two heads: $\{ HHT, HTH, THH \}$ Outcomes with exactly three heads: $\{ HHH \}$ There are $3 + 1 = 4$ favorable outcomes.

So, the probability $P(\text{at least two heads})$ is: $P(\text{at least two heads}) = \frac{4}{8} = \frac{1}{2}$

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Final Answers:

(i) Probability of getting exactly one head: $\frac{3}{8}$
(ii) Probability of getting at least two heads: $\frac{1}{2}$

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Would you like further explanation on these results or related concepts?

Here are 5 questions for further exploration:

1. What is the probability of getting no heads in three tosses?
2. How does the probability change if we toss the coin four times?
3. What is the probability of getting exactly two tails?
4. How can we generalize this result for $n$ tosses?
5. What is the probability of getting exactly three heads in three tosses?

Tip: For problems involving multiple coin tosses, visualize outcomes using a probability tree or list the outcomes directly to ensure accuracy.