Let's analyze the problem step-by-step.

Total Outcomes

When tossing a coin 3 times, there are $2^3 = 8$ possible outcomes. These outcomes are:

• TTT, HTH, TTH, HHT, HHH, THH, THT, HTT

Each outcome is equally likely, with a probability of $\frac{1}{8}$.

Event A: Two or more tails

Here, we select outcomes with at least two tails.

• TTT → 3 tails
• TTH → 2 tails
• THT → 2 tails
• HTT → 2 tails

Thus, the outcomes are TTT, TTH, THT, HTT.
The probability is:
$P(\text{Event A}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{8} = \frac{1}{2}.$

Event B: A tail on the first toss or the third toss (or both)

Here, we select outcomes where:

1. The first toss is T, or
2. The third toss is T, or both.

From the outcomes:

• First toss = T → TTT, TTH, THT, THH
• Third toss = T → TTT, HTH, THT, HTT

Now combine these outcomes (avoid duplicates):

• TTT, TTH, THT, THH, HTH, HTT

Thus, the outcomes are TTT, TTH, THT, THH, HTH, HTT.
The probability is:
$P(\text{Event B}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{6}{8} = \frac{3}{4}.$

Event C: A head on each of the last two tosses

Here, the last two tosses must be H.

• Only outcomes where the last two positions are H are valid:
  HHH, THH

Thus, the outcomes are HHH, THH.
The probability is:
$P(\text{Event C}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{2}{8} = \frac{1}{4}.$

Final Probabilities:

1. Event A: $\frac{1}{2}$
2. Event B: $\frac{3}{4}$
3. Event C: $\frac{1}{4}$

Let me know if you need further explanations!

Related Questions:

1. What is the complement of Event A?
2. How many outcomes have exactly one tail?
3. What is the probability of getting at least one head in 3 tosses?
4. How many outcomes satisfy both Event A and Event B?
5. What is the probability of getting no tails?

Tip: Always list all possible outcomes systematically when solving probability problems to avoid missing cases.