Question

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.

Explanation:

Step1: Determine the number of total outcomes

Since a coin is tossed 3 times, and each toss has 2 possible outcomes (head or tail), the total number of outcomes \(n = 2\times2\times2=8\). The probability of each single - outcome \(p=\frac{1}{8}\)

Step2: Analyze Event A (Two or more heads)

Count the number of outcomes with two or more heads:

• \(THH\) (2 heads), \(HTH\) (2 heads), \(HHT\) (2 heads), \(HHH\) (3 heads). So \(n(A)=4\)

The probability \(P(A)=\frac{n(A)}{n}=\frac{4}{8}=\frac{1}{2}\)

Step3: Analyze Event B (Exactly one head)

Count the number of outcomes with exactly one head:

• \(THT\) (1 head), \(HTT\) (1 head), \(TTH\) (1 head). So \(n(B) = 3\)

The probability \(P(B)=\frac{n(B)}{n}=\frac{3}{8}\)

Step4: Analyze Event C (A head on each of the first two tosses)

Count the number of outcomes with heads on the first two tosses:

• \(HHT\) (first two heads), \(HHH\) (first two heads). So \(n(C)=2\)

The probability \(P(C)=\frac{n(C)}{n}=\frac{2}{8}=\frac{1}{4}\)

Answer:

For Event A: Probability is \(\frac{1}{2}\)
For Event B: Probability is \(\frac{3}{8}\)
For Event C: Probability is \(\frac{1}{4}\)