In a problem where a coin is tossed three times, there are $2^3 = 8$ possible outcomes, because each toss has two outcomes: head (H) or tail (T).

The possible outcomes are:

$\{ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT \}$

Now, let's solve each part of the problem.

(i) Probability of getting exactly one head:

The outcomes with exactly one head are:

$\{ HTT, THT, TTH \}$

There are 3 such outcomes. Since there are 8 possible outcomes in total, the probability of getting exactly one head is:

$P(\text{exactly one head}) = \frac{3}{8}$

(ii) Probability of getting at least two heads:

To find the probability of getting at least two heads, we consider the outcomes where there are two or three heads. These outcomes are:

$\{ HHT, HTH, THH, HHH \}$

There are 4 such outcomes. Therefore, the probability of getting at least two heads is:

$P(\text{at least two heads}) = \frac{4}{8} = \frac{1}{2}$

Summary:

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

Would you like further details or have any other questions?

Related Questions:

1. What is the probability of getting no heads at all?
2. What is the probability of getting exactly two tails?
3. What is the probability of getting at least one head?
4. If the coin is biased and the probability of heads is $p$, how would this affect the outcomes?
5. How would the probabilities change if the coin was tossed 4 times?

Tip:

When calculating probabilities for multiple events, listing out all possible outcomes (like heads or tails for coin tosses) can help ensure you account for every case.