Answer
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Hint: A coin is tossed three times. The number of possible cases are the chances of getting head and tail in each move simultaneously. Here we find the number of possibilities by finding the sample space initially.
Complete step-by-step answer:
It is given that a coin is tossed three times.
The possible cases are either of getting head or tail.
The head and tail are defined as,
Head=H
Tail = T
The number of possible cases of the sample place is \[\left\{ {HHH,HHT,HTH,HTT,THH,TTH,THT,TTT} \right\}\]
The number of possible cases \[n\left( S \right) = 8\]
Also given that, A is the event that head appears once, B is the event that head appeared at the most twice, C is the event that the tail does not appear.
And we have to find the number of possible cases,
Let us consider event A: Head appears once.
The number of possible cases for event A is \[\left\{ {HTT,TTH,THT} \right\}\]
Here we consider the possibility having only one head in three tosses,
\[n\left( A \right) = 3\]
Where \[n\left( A \right) = 3\] is the number of possibilities of getting head only once.
Let us consider event B: Head appeared at the most twice.
At most twice implies there can be one head, two heads or no heads.
The number of possible cases for event A is \[\left\{ {HHT,HTH,HTT,THH,TTH,THT,TTT} \right\}\]
\[n\left( B \right) = 7\]
The number of possibilities of getting heads at most twice is \[n\left( B \right) = 7\].
Let us consider event C: Tail does not appear.
The number of possible cases for event C is \[\left\{ {HHH} \right\}\]
\[n\left( C \right) = 1\]
The number of possibilities of getting no tails is \[n\left( C \right) = 1\]
Hence,
\[n\left( A \right) = 3,{\rm{ }}n\left( B \right) = 7,n \left( C \right) = 1\].
Note: A population is the entire group that you want to draw conclusions about and a sample is the specific group that you will collect data from. The size of the sample is always less than the total size of the population.
Here the number of possibilities of getting no heads is equal to the number of possibilities of getting at least 3 heads.
Complete step-by-step answer:
It is given that a coin is tossed three times.
The possible cases are either of getting head or tail.
The head and tail are defined as,
Head=H
Tail = T
The number of possible cases of the sample place is \[\left\{ {HHH,HHT,HTH,HTT,THH,TTH,THT,TTT} \right\}\]
The number of possible cases \[n\left( S \right) = 8\]
Also given that, A is the event that head appears once, B is the event that head appeared at the most twice, C is the event that the tail does not appear.
And we have to find the number of possible cases,
Let us consider event A: Head appears once.
The number of possible cases for event A is \[\left\{ {HTT,TTH,THT} \right\}\]
Here we consider the possibility having only one head in three tosses,
\[n\left( A \right) = 3\]
Where \[n\left( A \right) = 3\] is the number of possibilities of getting head only once.
Let us consider event B: Head appeared at the most twice.
At most twice implies there can be one head, two heads or no heads.
The number of possible cases for event A is \[\left\{ {HHT,HTH,HTT,THH,TTH,THT,TTT} \right\}\]
\[n\left( B \right) = 7\]
The number of possibilities of getting heads at most twice is \[n\left( B \right) = 7\].
Let us consider event C: Tail does not appear.
The number of possible cases for event C is \[\left\{ {HHH} \right\}\]
\[n\left( C \right) = 1\]
The number of possibilities of getting no tails is \[n\left( C \right) = 1\]
Hence,
\[n\left( A \right) = 3,{\rm{ }}n\left( B \right) = 7,n \left( C \right) = 1\].
Note: A population is the entire group that you want to draw conclusions about and a sample is the specific group that you will collect data from. The size of the sample is always less than the total size of the population.
Here the number of possibilities of getting no heads is equal to the number of possibilities of getting at least 3 heads.
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