Answered

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.

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
& \multicolumn{8}{|c|}{Outcomes} & \multirow{2}{*}{Probability} \\
\hline
& THH & HHT & HTT & HHH & HTH & TTH & THT & TTT & \\
\hline
\begin{tabular}{l}
Event A: A tail on both the first and the \\
last tosses
\end{tabular}
& & & $\square$ & & & & $\square$ & $\square$ & \\
\hline
\begin{tabular}{l}
Event B: Alternating tail and head (with \\
either coming first)
\end{tabular}
& & & & & $\square$ & $\square$ & $\square$ & & \\
\hline
\begin{tabular}{l}
Event C: A head on each of the first two \\
tosses
\end{tabular}
& & $\square$ & & $\square$ & & & & & \\
\hline
\end{tabular}
\][/tex]



Answer :

GinnyAnswer
Alright, let's go through each event and determine the outcomes that meet the criteria for each event. Then we will compute the probability of each event.

### Event A: A tail on both the first and the last tosses
We need to find all outcomes where the first and last tosses are tails ('T').
The possible outcomes of three coin tosses are:

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

For Event A, the outcomes where the first and the last tosses are tails are:
```
THT, TTT
```

Since there are 8 equally likely outcomes, the probability of Event A is the number of favorable outcomes divided by the total number of outcomes:

- Favorable outcomes for Event A: `THT, TTT`
- Number of favorable outcomes: 2
- Total number of outcomes: 8

Therefore, the probability of Event A is:
[tex]\[ \frac{2}{8} = 0.25 \][/tex]

### Event B: Alternating tail and head (with either coming first)
We need to find all outcomes where tails and heads alternate.

The outcomes where tails and heads alternate are:
```
HTH, THT
```

For Event B, the outcomes are:
```
HTH, THT
```

Again, considering the probability computation:

- Favorable outcomes for Event B: `HTH, THT`
- Number of favorable outcomes: 2
- Total number of outcomes: 8

Hence, the probability of Event B is:
[tex]\[ \frac{2}{8} = 0.25 \][/tex]

### Event C: A head on each of the first two tosses
We need to find all outcomes where the first and second tosses are heads ('H').

The outcomes where the first two tosses are heads are:
```
HHH, HHT
```

For Event C, the outcomes are:
```
HHH, HHT
```

For probability:

- Favorable outcomes for Event C: `HHH, HHT`
- Number of favorable outcomes: 2
- Total number of outcomes: 8

Thus, the probability of Event C is:
[tex]\[ \frac{2}{8} = 0.25 \][/tex]

### Summary Table

Combining all the collected information in a structured table format:

[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline & HHH & HHT & HTH & HTT & THH & THT & TTH & TTT & \text{Probability} \\ \hline \begin{array}{l} \text{Event A: A tail on both the} \\ \text{first and the last tosses} \end{array} & & & & & & \checkmark & & \checkmark & 0.25 \\ \hline \begin{array}{l} \text{Event B: Alternating tail and head} \\ \text{(with either coming first)} \end{array} & & & \checkmark & & & \checkmark & & & 0.25 \\ \hline \begin{array}{l} \text{Event C: A head on each of the} \\ \text{first two tosses} \end{array} & \checkmark & \checkmark & & & & & & & 0.25 \\ \hline \end{array} \][/tex]

This table indicates which outcomes belong to each event and the corresponding probabilities.

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