A coin is tossed three times. An outcome is represented by a string such as 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.

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



Answer :

Sure! Let's go through each event step by step and identify the corresponding outcomes. Then, we'll determine the probabilities based on the given data.

### Event A: A tail on both the first and last tosses
We need outcomes where the first toss is T (tail) and the third toss is T (tail). Looking at the outcomes list:
- THT (Tail, Head, Tail)
- TTT (Tail, Tail, Tail)

There are 2 such outcomes out of 8 possible outcomes.

Probability of Event A:
Number of favorable outcomes = 2
Total number of outcomes = 8
[tex]\[ \text{Probability} = \frac{2}{8} = 0.25 \][/tex]

### Event B: A head on each of the last two tosses
We need outcomes where the second and third tosses are both H (head). Looking at the outcomes list:
- THH (Tail, Head, Head)
- HHH (Head, Head, Head)
- HHT (Head, Head, Tail)

There are 3 such outcomes out of 8 possible outcomes.

Probability of Event B:
Number of favorable outcomes = 3
Total number of outcomes = 8
[tex]\[ \text{Probability} = \frac{3}{8} = 0.375 \][/tex]

### Event C: Alternating tail and head (with either coming first)
We need outcomes where the tosses alternate between T (tail) and H (head):
- HTH (Head, Tail, Head)
- THT (Tail, Head, Tail)

There are 2 such outcomes out of 8 possible outcomes.

Probability of Event C:
Number of favorable outcomes = 2
Total number of outcomes = 8
[tex]\[ \text{Probability} = \frac{2}{8} = 0.25 \][/tex]

### Summary
Let's summarize these probabilities in the given table format:

[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \text{} & \multicolumn{8}{|c|}{ \text{Outcomes} } & \multirow{2}{*}{ \text{Probability} } \\ \hline \text{} & \text{HHH} & \text{HHT} & \text{HTH} & \text{HTT} & \text{THH} & \text{THT} & \text{TTH} & \text{TTT} & \\ \hline \begin{array}{l} \text{Event A: A tail on both the first and the} \\ \text{last tosses} \end{array} & & & & & & \checkmark & & \checkmark & 0.25 \\ \hline \begin{array}{l} \text{Event B: A head on each of the last two} \\ \text{tosses} \end{array} & \checkmark & \checkmark & & & \checkmark & & & & 0.375 \\ \hline \begin{array}{l} \text{Event C: Alternating tail and head (with} \\ \text{either coming first)} \end{array} & & & \checkmark & & & \checkmark & & & 0.25 \\ \hline \end{array} \][/tex]

Thus, the probabilities for Event A, Event B, and Event C are [tex]\(0.25\)[/tex], [tex]\(0.375\)[/tex], and [tex]\(0.25\)[/tex] respectively.