Let's analyze the table you've provided. We're dealing with a scenario where a coin is tossed three times. The outcomes and the events are listed, and our goal is to identify which outcomes belong to each event and then calculate the probability for each event.

Outcomes:

The 8 possible outcomes for three coin tosses are:

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

Each of these outcomes has a probability of $\frac{1}{8}$, because each outcome is equally likely.

Events:

• Event A: A tail on the last toss.
• Event B: More heads than tails.
• Event C: A tail on the last toss or the second toss (or both).

Checking Outcomes for Each Event:

1. Event A: A tail on the last toss

   • The outcomes with tails on the last toss are: HHT, HTT, THT, TTT.
   • So, Event A includes these outcomes: HHT, HTT, THT, TTT.

2. Event B: More heads than tails

   • The outcomes with more heads than tails are: HHH, HHT, HTH, THH.
   • So, Event B includes these outcomes: HHH, HHT, HTH, THH.

3. Event C: A tail on the last toss or the second toss (or both)

   • The outcomes with a tail on the last toss or the second toss are: HHT, HTT, THT, TTH, TTT.
   • So, Event C includes these outcomes: HHT, HTT, THT, TTH, TTT.

Calculating the Probability for Each Event:

The probability of an event is calculated as: $P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{\text{Number of outcomes in the event}}{8}$

1. Probability of Event A:

   • Number of outcomes: 4 (HHT, HTT, THT, TTT)
   • $P(A) = \frac{4}{8} = 0.5$

2. Probability of Event B:

   • Number of outcomes: 4 (HHH, HHT, HTH, THH)
   • $P(B) = \frac{4}{8} = 0.5$

3. Probability of Event C:

   • Number of outcomes: 5 (HHT, HTT, THT, TTH, TTT)
   • $P(C) = \frac{5}{8} = 0.625$

Summary:

• Event A (A tail on the last toss): Outcomes = HHT, HTT, THT, TTT; Probability = 0.5
• Event B (More heads than tails): Outcomes = HHH, HHT, HTH, THH; Probability = 0.5
• Event C (A tail on the last or second toss): Outcomes = HHT, HTT, THT, TTH, TTT; Probability = 0.625

Would you like to explore more details, or do you have any further questions?

Here are some related questions you might explore:

1. What is the probability of getting exactly two heads in three tosses?
2. What is the probability of getting at least one head in three tosses?
3. If the coin was biased with a probability of 0.6 for heads, how would the probabilities change?
4. What is the probability of getting the same result (all heads or all tails) in three tosses?
5. How do you compute the expected number of heads in three tosses?
6. What is the probability of getting exactly one tail in three tosses?
7. How would the outcome probabilities change if we tossed the coin four times?
8. What is the probability that the first and last tosses result in heads?

Tip: When working with multiple events, it's helpful to visualize the sample space and use set operations (like unions and intersections) to combine events effectively.