Let's analyze the events based on the outcomes given:

1. Event A: No heads on the last two tosses
   This event means that the last two tosses must be tails (TT). The possible outcomes for this event are:

   • TTT and HTT

2. Event B: A tail on the last toss
   This event means that the final toss must be a tail (T). The possible outcomes for this event are:

   • TTT, HTT, THT, HHT

3. Event C: A head on each of the last two tosses
   This event requires both of the last two tosses to be heads (HH). The possible outcomes for this event are:

   • THH and HHH

Now let's determine the probabilities. Since there are 8 possible outcomes and each is equally likely, the probability of each outcome is $\frac{1}{8}$.

• Probability of Event A: 2 outcomes (TTT, HTT) $\Rightarrow \frac{2}{8} = \frac{1}{4}$
• Probability of Event B: 4 outcomes (TTT, HTT, THT, HHT) $\Rightarrow \frac{4}{8} = \frac{1}{2}$
• Probability of Event C: 2 outcomes (THH, HHH) $\Rightarrow \frac{2}{8} = \frac{1}{4}$

If you want more detailed explanations or have any questions, let me know.

Related Questions

1. What is the probability of getting exactly two heads in three coin tosses?
2. How would the probabilities change if the coin were biased?
3. What if the coin was tossed four times—how would Event A change?
4. How can we calculate the probability of getting at least one head in three tosses?
5. What is the probability of getting heads in all three tosses?

Tip: In probability problems, always start by identifying the total number of possible outcomes, then match each event's conditions to these outcomes.