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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.

\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|}
\cline { 2 - 10 }
\multicolumn{1}{c|}{} & \multicolumn{8}{c|}{Outcomes} & \\
\hline
& TTH & HHT & HHH & HTT & THT & HTH & THH & THT & Probability \\
\hline
Event A: Exactly one head & & & & & & & & & \\
\hline
Event B: More heads than tails & & & & & & & & & \\
\hline
\begin{tabular}{l}
Event C: A head on each of the first two \\
tosses
\end{tabular}
& & & & & & & & & \\
\hline
\end{tabular}



Answer :

GinnyAnswer
Let's solve each part of the question step-by-step.

### Event A: Exactly one head
We need to check which outcomes have exactly one head.

- TTH: This outcome has 1 head.
- HHT: This outcome has 2 heads.
- HHH: This outcome has 3 heads.
- HTT: This outcome has 1 head.
- THT: This outcome has 1 head.
- HTH: This outcome has 2 heads.
- THH: This outcome has 2 heads.
- THT: This outcome has 1 head.

So, the outcomes for Event A are:
- TTH, HTT, THT, THT

Now, we count the number of outcomes and calculate the probability. There are 4 outcomes that satisfy Event A out of the total 8 outcomes.

Probability of Event A = [tex]\( \frac{4}{8} = 0.5 \)[/tex]

### Event B: More heads than tails
We need to check which outcomes have more heads than tails.

- TTH: 1 head, 2 tails.
- HHT: 2 heads, 1 tail.
- HHH: 3 heads, 0 tails.
- HTT: 1 head, 2 tails.
- THT: 1 head, 2 tails.
- HTH: 2 heads, 1 tail.
- THH: 2 heads, 1 tail.
- THT: 1 head, 2 tails.

So, the outcomes for Event B are:
- HHT, HHH, HTH, THH

Now, we count the number of outcomes and calculate the probability. There are 4 outcomes that satisfy Event B out of the total 8 outcomes.

Probability of Event B = [tex]\( \frac{4}{8} = 0.5 \)[/tex]

### Event C: A head on each of the first two tosses
We need to check which outcomes start with HH.

- TTH: Does not start with HH.
- HHT: Starts with HH.
- HHH: Starts with HH.
- HTT: Does not start with HH.
- THT: Does not start with HH.
- HTH: Starts with H but the second toss is not H.
- THH: Does not start with HH.
- THT: Does not start with HH.

So, the outcomes for Event C are:
- HHT, HHH

Now, we count the number of outcomes and calculate the probability. There are 2 outcomes that satisfy Event C out of the total 8 outcomes.

Probability of Event C = [tex]\( \frac{2}{8} = 0.25 \)[/tex]

### Summary Table:
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|}
\hline & TTH & HHT & HHH & HTT & THT & HTH & THH & THT & Probability \\
\hline Event A: Exactly one head & X & & & X & X & & & X & 0.5 \\
\hline Event B: More heads than tails & & X & X & & & X & X & & 0.5 \\
\hline \begin{tabular}{l}
Event C: A head on each of the first two \\
tosses \end{tabular} & & X & X & & & & & & 0.25 \\
\hline
\end{tabular}

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