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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}{|c|c|c|c|c|c|c|c|c|c|}
\hline
& \multicolumn{8}{|c|}{Outcomes} & \multirow{2}{*}{Probability} \\
\hline
& HHH & HHT & HTH & HTT & THH & THT & TTH & TTT & \\
\hline
Event A: Two or more tails & [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
\begin{tabular}{l}
Event B: A head on each of the last two \\
tosses
\end{tabular} & [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: A tail on the last toss & [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 :

GinnyAnswer
To solve the problem of finding the probability for each given event when a coin is tossed three times, we should carefully analyze each event and determine the number of favorable outcomes. Here is the detailed step-by-step solution:

### Event A: Two or more tails

1. Identify outcomes with two or more tails:
- HHH: 0 tails
- HTH: 1 tail
- HHT: 1 tail
- HTT: 2 tails (favorable)
- TTH: 2 tails (favorable)
- THT: 2 tails (favorable)
- TTT: 3 tails (favorable)
- THH: 1 tail

2. Favorable outcomes for Event A:
HTT, TTH, THT, TTT

3. Number of favorable outcomes for Event A: 4

4. Probability of Event A:
Since there are 8 possible outcomes and 4 favorable ones:
[tex]\( P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{8} = 0.5 \)[/tex]

### Event B: A head on each of the last two tosses

1. Identify outcomes with a head on each of the last two tosses:
- HHH: last two tosses are HH (favorable)
- HTH: last two tosses are TH
- HHT: last two tosses are HT
- HTT: last two tosses are TT
- TTH: last two tosses are TH
- THT: last two tosses are HT
- TTT: last two tosses are TT
- THH: last two tosses are HH (favorable)

2. Favorable outcomes for Event B:
HHH, THH

3. Number of favorable outcomes for Event B: 2

4. Probability of Event B:
Since there are 8 possible outcomes and 2 favorable ones:
[tex]\( P(B) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{8} = 0.25 \)[/tex]

### Event C: A tail on the last toss

1. Identify outcomes with a tail on the last toss:
- HHH: last toss is H
- HTH: last toss is H
- HHT: last toss is T (favorable)
- HTT: last toss is T (favorable)
- TTH: last toss is H
- THT: last toss is T (favorable)
- TTT: last toss is T (favorable)
- THH: last toss is H

2. Favorable outcomes for Event C:
HHT, HTT, THT, TTT

3. Number of favorable outcomes for Event C: 4

4. Probability of Event C:
Since there are 8 possible outcomes and 4 favorable ones:
[tex]\( P(C) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{8} = 0.5 \)[/tex]

### Summary:

| Event | Favorable Outcomes | Number of Favorable Outcomes | Probability |
|-------|---------------------|------------------------------|-------------|
| A | HTT, TTH, THT, TTT | 4 | 0.5 |
| B | HHH, THH | 2 | 0.25 |
| C | HHT, HTT, THT, TTT | 4 | 0.5 |

Thus, the probabilities for Events A, B, and C are 0.5, 0.25, and 0.5, respectively.

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