Question

A coin is tossed three times. An outcome is represented by a string of the sort HTT (meaning a head on the first toss, followed by two tails). The 8 outcomes are listed in the table below. Note that each outcome has the same probability.

For each of the three events in the table, check the outcome(s) that are contained in the event. Then, in the last column, enter the probability of the event.

Outcomes Probability
TTT HTT HHH TTH HTH HHT THH THT
Event A: Exactly one head
Event B: No tails on the last two tosses
Event C: Alternating tail and head (with either coming first)

Answer 1

Let's analyze the outcomes first. There are 8 possible outcomes when tossing a coin three times:

1. TTT
2. HTT
3. HHH
4. TTH
5. HTH
6. HHT
7. THH
8. THT

Now, we'll check the given events and the outcomes contained within them, as well as calculate their probabilities.

### Event A: Exactly one head
The outcomes that have exactly one head are:
- HTT
- TTH
- THT

So, the outcomes contained in Event A are: **HTT, TTH, THT**.

**Probability of Event A**:
There are 3 outcomes with exactly one head out of 8 total outcomes.
\[
P(A) = \frac{3}{8}
\]

### Event B: No tails on the last two tosses
The outcomes with no tails on the last two tosses (meaning the last two must be heads) are:
- HHH
- HHT
- HTH

So, the outcomes contained in Event B are: **HHH, HHT, HTH**.

**Probability of Event B**:
There are 3 outcomes with no tails on the last two tosses out of 8 total outcomes.
\[
P(B) = \frac{3}{8}
\]

### Event C: Alternating tail and head (with either coming first)
The outcomes that alternate between heads and tails are:
- HTH
- THT

So, the outcomes contained in Event C are: **HTH, THT**.

**Probability of Event C**:
There are 2 outcomes that alternate between heads and tails out of 8 total outcomes.
\[
P(C) = \frac{2}{8} = \frac{1}{4}
\]

### Summary Table
Now, let's summarize this information:

\[
\begin{array}{|c|c|c|}
\hline
\text{Outcomes} & \text{Contains in Event} & \text{Probability} \\
\hline
TTT & & \\
HTT & \text{Yes (A)} & \\
HHH & \text{Yes (B)} & \\
TTH & \text{Yes (A)} & \\
HTH & \text{Yes (B/C)} & \\
HHT & \text{Yes (B)} & \\
THH & & \\
THT & \text{Yes (A/C)} & \\
\hline
\text{Event A: Exactly one head} & \text{HTT, TTH, THT} & \frac{3}{8} \\
\text{Event B: No tails on the last two tosses} & \text{HHH, HHT, HTH} & \frac{3}{8} \\
\text{Event C: Alternating tail and head} & \text{HTH, THT} & \frac{1}{4} \\
\hline
\end{array}
\]

So the final probabilities are:
- Event A: \(\frac{3}{8}\)
- Event B: \(\frac{3}{8}\)
- Event C: \(\frac{1}{4}\)