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 second toss or the third toss (or both) event c: a tail on the second toss

Explanation:

Step1: Recall probability formula

The probability of an event $P(E)=\frac{n(E)}{n(S)}$, where $n(E)$ is the number of elements in the event $E$ and $n(S)$ is the number of elements in the sample - space. Here, $n(S) = 8$ since there are 8 possible outcomes when a coin is tossed 3 times.

Step2: Analyze Event A

Event A: No tails on the first two tosses. The favorable outcomes are HHH and HHT. So $n(A)=2$. Then $P(A)=\frac{n(A)}{n(S)}=\frac{2}{8}=\frac{1}{4}$.

Step3: Analyze Event B

Event B: A tail on the second toss or the third toss (or both). The non - favorable outcome is HHH. So $n(B)=7$. Then $P(B)=\frac{n(B)}{n(S)}=\frac{7}{8}$.

Step4: Analyze Event C

Event C: A tail on the second toss. The favorable outcomes are TTH, THT, HTH, HTT. So $n(C)=4$. Then $P(C)=\frac{n(C)}{n(S)}=\frac{4}{8}=\frac{1}{2}$.

Answer:

Event A: Check HHH, HHT; Probability: $\frac{1}{4}$
Event B: Check TTH, THT, HTH, HTT, HHT, THH, TTT; Probability: $\frac{7}{8}$
Event C: Check TTH, THT, HTH, HTT; Probability: $\frac{1}{2}$