Questions: Counting and Probability Outcomes and event probability 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 --- --- --- --- --- --- --- --- --- --- TH HHT THT HTH HTT THH T1T HHH Event A: Exactly one head — 0 0 — Event B: A tail on the last toss ■ (I) Event C: More tails than heads ■ ■ ■ D (T) —

Counting and Probability Outcomes and event probability 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 --- --- --- --- --- --- --- --- --- --- TH HHT THT HTH HTT THH T1T HHH Event A: Exactly one head — 0 0 — Event B: A tail on the last toss ■ (I) Event C: More tails than heads ■ ■ ■ D (T) —
Transcript text: Counting and Probability Outcomes and event probability 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 & TH & HHT & THT & HTH & HTT & THH & T1T & HHH & \\ \hline Event A: Exactly one head & $\square$ & — & $\square$ & $\square$ & $\square$ & $\square$ & 0 & 0 & — \\ \hline Event B: A tail on the last toss & $\square$ & $\square$ & $\square$ & $\square$ & $\square$ & ■ & $\square$ & (I) & $\square$ \\ \hline Event C: More tails than heads & $\square$ & $\square$ & ■ & ■ & ■ & D & (T) & $\square$ & — \\ \hline \end{tabular}
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Solution

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Solution Steps

Solution Approach
  1. Identify the outcomes for each event based on the given conditions.
  2. Calculate the probability of each event by dividing the number of favorable outcomes by the total number of possible outcomes (which is 8 for three coin tosses).
Step 1: Identify Outcomes for Event A

For Event A, which requires exactly one head, the favorable outcomes are:

  • \( \text{TH} \)
  • \( \text{THT} \)
  • \( \text{HTT} \)

Thus, the number of favorable outcomes is 3.

Step 2: Calculate Probability for Event A

The probability of Event A is calculated as: \[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{3}{8} = 0.375 \]

Step 3: Identify Outcomes for Event B

For Event B, which requires a tail on the last toss, the favorable outcomes are:

  • \( \text{HHT} \)
  • \( \text{THT} \)
  • \( \text{HTT} \)
  • \( \text{T1T} \)

Thus, the number of favorable outcomes is 4.

Step 4: Calculate Probability for Event B

The probability of Event B is calculated as: \[ P(B) = \frac{4}{8} = 0.5 \]

Step 5: Identify Outcomes for Event C

For Event C, which requires more tails than heads, the favorable outcomes are:

  • \( \text{THT} \)
  • \( \text{HTT} \)
  • \( \text{T1T} \)

Thus, the number of favorable outcomes is 3.

Step 6: Calculate Probability for Event C

The probability of Event C is calculated as: \[ P(C) = \frac{3}{8} = 0.375 \]

Final Answer

The probabilities for the events are:

  • \( P(A) = 0.375 \)
  • \( P(B) = 0.5 \)
  • \( P(C) = 0.375 \)

Thus, the final answers are: \[ \boxed{P(A) = 0.375, \, P(B) = 0.5, \, P(C) = 0.375} \]

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