Questions: Outcomes are not probability Event A: Alternating hats and tails (with either coming first) Event B: A head on each of the first two tosses Event C: Two or more heads TTH THH HTT HTH HHT HHH TTT THT

Solution Steps

To calculate the probability of each event, we need to determine the number of favorable outcomes for each event and divide it by the total number of possible outcomes. The total number of possible outcomes is the length of the list provided. For each event, count the number of outcomes that satisfy the event's condition.

The total number of possible outcomes when tossing three coins is given by the length of the outcomes list: \[ \text{Total Outcomes} = 8 \]

Event A consists of outcomes that alternate hats and tails, which are \( \{HTH, THT\} \). The number of favorable outcomes is: \[ \text{Favorable Outcomes for A} = 2 \] Thus, the probability of Event A is calculated as: \[ P(A) = \frac{\text{Favorable Outcomes for A}}{\text{Total Outcomes}} = \frac{2}{8} = \frac{1}{4} = 0.25 \]

Event B consists of outcomes where there is a head on each of the first two tosses, which are \( \{HHT, HHH\} \). The number of favorable outcomes is: \[ \text{Favorable Outcomes for B} = 2 \] Thus, the probability of Event B is: \[ P(B) = \frac{\text{Favorable Outcomes for B}}{\text{Total Outcomes}} = \frac{2}{8} = \frac{1}{4} = 0.25 \]

Event C consists of outcomes with two or more heads, which are \( \{THH, HTH, HHT, HHH\} \). The number of favorable outcomes is: \[ \text{Favorable Outcomes for C} = 4 \] Thus, the probability of Event C is: \[ P(C) = \frac{\text{Favorable Outcomes for C}}{\text{Total Outcomes}} = \frac{4}{8} = \frac{1}{2} = 0.5 \]

Final Answer