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 tail on both the first and the last tosses event b: a head on each of the first two tosses event c: a tail on the first toss or the third toss (or both)

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 tail on both the first and the last tosses event b: a head on each of the first two tosses event c: a tail on the first toss or the third toss (or both)

Answer

Explanation:

Step1: Determine total number of outcomes

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

Step2: Analyze Event A

Event A: A tail on both the first and the last tosses. The favorable outcomes are TTH and TTT. 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: A head on each of the first two tosses. The favorable outcomes are HHH and HHT. So the number of favorable outcomes $n(B)=2$. The probability $P(B)=\frac{n(B)}{n(S)}=\frac{2}{8}=\frac{1}{4}$.

Step4: Analyze Event C

Event C: A tail on the first toss or the third toss (or both). The favorable outcomes are TTT, TTH, THT, THH, HTT, HTH. So the number of favorable outcomes $n(C)=6$. The probability $P(C)=\frac{n(C)}{n(S)}=\frac{6}{8}=\frac{3}{4}$.

Answer:

Event Outcomes Probability
A TTH, TTT $\frac{1}{4}$
B HHH, HHT $\frac{1}{4}$
C TTT, TTH, THT, THH, HTT, HTH $\frac{3}{4}$