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. event a: no tails on the first two tosses event b: a tail on the first toss event c: alternating tail and head (with either coming first)

Explanation:

Step1: Determine the total number of outcomes

When a coin is tossed three times, each toss has 2 possible outcomes (head or tail). So the total number of outcomes is \(2\times2\times2 = 8\). Each outcome has a probability of \(\frac{1}{8}\)

Step2: Analyze Event A: No tails on the first two tosses

For no tails on the first two tosses, the first two tosses must be heads. The third toss can be either head or tail. So the outcomes are \(HHH\) and \(HHT\). The number of favorable outcomes is 2. Probability \(P(A)=\frac{2}{8}=\frac{1}{4}\)

Step3: Analyze Event B: A tail on the first toss

If there is a tail on the first toss, the first - toss outcome is T. The second and third tosses can be either head or tail. The outcomes are \(T HH\), \(T HT\), \(T TH\), \(T TT\). The number of favorable outcomes is 4. Probability \(P(B)=\frac{4}{8}=\frac{1}{2}\)

Step4: Analyze Event C: Alternating tail and head (with either coming first)

Case 1: Tail first: \(T HT\)
Case 2: Head first: \(HTH\)
The number of favorable outcomes is 2. Probability \(P(C)=\frac{2}{8}=\frac{1}{4}\)

Answer:

For \(HTT\): \(A:0\), \(B:1\), \(C:0\)
For \(TTH\): \(A:0\), \(B:1\), \(C:0\)
For \(HHT\): \(A:1\), \(B:0\), \(C:0\)
For \(HHH\): \(A:1\), \(B:0\), \(C:0\)
For \(THH\): \(A:0\), \(B:1\), \(C:0\)
For \(THT\): \(A:0\), \(B:1\), \(C:1\)
For \(HTH\): \(A:0\), \(B:0\), \(C:1\)
For \(TTT\): \(A:0\), \(B:1\), \(C:0\)
Probability for each outcome: \(\frac{1}{8}\)