a coin is tossed three times. an outcome is represented by a string of the sort htt (meaning a head on the…

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: a head on each of the last two tosses event b: more tails than heads event c: no tails on the last two tosses

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: a head on each of the last two tosses event b: more tails than heads event c: no tails on the last two tosses

Answer

Explanation:

Step1: Determine total number of outcomes

There are 8 total outcomes when a coin is tossed 3 - times.

Step2: Analyze Event A

Event A: A head on each of the last two tosses. The favorable outcomes are HHH, HTH. So the number of favorable outcomes $n(A)=2$. The probability $P(A)=\frac{n(A)}{n(S)}=\frac{2}{8}=\frac{1}{4}$.

Step3: Analyze Event B

Event B: More tails than heads. The favorable outcomes are TTH, THT, HTT, TTT. So the number of favorable outcomes $n(B) = 4$. The probability $P(B)=\frac{n(B)}{n(S)}=\frac{4}{8}=\frac{1}{2}$.

Step4: Analyze Event C

Event C: No tails on the last two tosses. The favorable outcomes are HHH, THH. So the number of favorable outcomes $n(C)=2$. The probability $P(C)=\frac{n(C)}{n(S)}=\frac{2}{8}=\frac{1}{4}$.

Event Outcomes (Checked) Probability
Event A HHH, HTH $\frac{1}{4}$
Event B TTH, THT, HTT, TTT $\frac{1}{2}$
Event C HHH, THH $\frac{1}{4}$

Answer:

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