
Two dice are thrown. Describe the sample space of this experiment.
Answer
558.9k+ views
Hint:
1) If two dice are thrown, there are 6 × 6 = 36 different outcomes possible.
2) The sample space of a random experiment is the set of all possible outcomes.
3) The sample space is represented using S.
4) A subset of the sample space of an experiment is called an event represented by E.
Complete step by step solution:
When two dice are thrown, we may get an outcome as (1, 1), (2, 5), (1, 6), (3, 1) etc.
Since, there are six different possible outcomes for a dice, the set (S) of all the outcomes can be listed as follows:
\[\left( {1,{\text{ }}1} \right),{\text{ }}\left( {1,{\text{ }}2} \right),{\text{ }}\left( {1,{\text{ }}3} \right),{\text{ }}\left( {1,{\text{ }}4} \right),{\text{ }}\left( {1,{\text{ }}5} \right),{\text{ }}\left( {1,{\text{ }}6} \right){\text{ }} = {\text{ }}6{\text{ }}possibilities.\]
\[\left( {2,{\text{ }}1} \right),{\text{ }}\left( {2,{\text{ }}2} \right),{\text{ }}\left( {2,{\text{ }}3} \right),{\text{ }}\left( {2,{\text{ }}4} \right),{\text{ }}\left( {2,{\text{ }}5} \right),{\text{ }}\left( {2,{\text{ }}6} \right){\text{ }} = {\text{ }}6{\text{ }}possibilities.\]
\[\left( {3,{\text{ }}1} \right),{\text{ }}\left( {3,{\text{ }}2} \right),{\text{ }}\left( {3,{\text{ }}3} \right),{\text{ }}\left( {3,{\text{ }}4} \right),{\text{ }}\left( {3,{\text{ }}5} \right),{\text{ }}\left( {3,{\text{ }}6} \right){\text{ }} = {\text{ }}6{\text{ }}possibilities.\]
\[\;\left( {4,{\text{ }}1} \right),{\text{ }}\left( {4,{\text{ }}2} \right),{\text{ }}\left( {4,{\text{ }}3} \right),{\text{ }}\left( {4,{\text{ }}4} \right),{\text{ }}\left( {4,{\text{ }}5} \right),{\text{ }}\left( {4,{\text{ }}6} \right){\text{ }} = {\text{ }}6{\text{ }}possibilities.\]
\[\;\left( {5,{\text{ }}1} \right),{\text{ }}\left( {5,{\text{ }}2} \right),{\text{ }}\left( {5,{\text{ }}3} \right),{\text{ }}\left( {5,{\text{ }}4} \right),{\text{ }}\left( {5,{\text{ }}5} \right),{\text{ }}\left( {5,{\text{ }}6} \right){\text{ }} = {\text{ }}6{\text{ }}possibilities.\]
\[\;\left( {6,{\text{ }}1} \right),{\text{ }}\left( {6,{\text{ }}2} \right),{\text{ }}\left( {6,{\text{ }}3} \right),{\text{ }}\left( {6,{\text{ }}4} \right),{\text{ }}\left( {6,{\text{ }}5} \right),{\text{ }}\left( {6,{\text{ }}6} \right){\text{ }} = {\text{ }}6{\text{ }}possibilities.\]
Total number of elements (possibilities) of set S are therefore,\[n\left( S \right) = 6 \times 6 = 36\]; i.e. six possibilities of second dice for each of the six possibilities of the first dice.
Note:
1) A sample space is usually denoted using set notation, and the possible ordered outcomes are listed as elements in the set.
2) The probability of an outcome E in a sample space S is a number P between 1 and 0 that measures the likelihood that E will occur on a single trial.
1) If two dice are thrown, there are 6 × 6 = 36 different outcomes possible.
2) The sample space of a random experiment is the set of all possible outcomes.
3) The sample space is represented using S.
4) A subset of the sample space of an experiment is called an event represented by E.
Complete step by step solution:
When two dice are thrown, we may get an outcome as (1, 1), (2, 5), (1, 6), (3, 1) etc.
Since, there are six different possible outcomes for a dice, the set (S) of all the outcomes can be listed as follows:
\[\left( {1,{\text{ }}1} \right),{\text{ }}\left( {1,{\text{ }}2} \right),{\text{ }}\left( {1,{\text{ }}3} \right),{\text{ }}\left( {1,{\text{ }}4} \right),{\text{ }}\left( {1,{\text{ }}5} \right),{\text{ }}\left( {1,{\text{ }}6} \right){\text{ }} = {\text{ }}6{\text{ }}possibilities.\]
\[\left( {2,{\text{ }}1} \right),{\text{ }}\left( {2,{\text{ }}2} \right),{\text{ }}\left( {2,{\text{ }}3} \right),{\text{ }}\left( {2,{\text{ }}4} \right),{\text{ }}\left( {2,{\text{ }}5} \right),{\text{ }}\left( {2,{\text{ }}6} \right){\text{ }} = {\text{ }}6{\text{ }}possibilities.\]
\[\left( {3,{\text{ }}1} \right),{\text{ }}\left( {3,{\text{ }}2} \right),{\text{ }}\left( {3,{\text{ }}3} \right),{\text{ }}\left( {3,{\text{ }}4} \right),{\text{ }}\left( {3,{\text{ }}5} \right),{\text{ }}\left( {3,{\text{ }}6} \right){\text{ }} = {\text{ }}6{\text{ }}possibilities.\]
\[\;\left( {4,{\text{ }}1} \right),{\text{ }}\left( {4,{\text{ }}2} \right),{\text{ }}\left( {4,{\text{ }}3} \right),{\text{ }}\left( {4,{\text{ }}4} \right),{\text{ }}\left( {4,{\text{ }}5} \right),{\text{ }}\left( {4,{\text{ }}6} \right){\text{ }} = {\text{ }}6{\text{ }}possibilities.\]
\[\;\left( {5,{\text{ }}1} \right),{\text{ }}\left( {5,{\text{ }}2} \right),{\text{ }}\left( {5,{\text{ }}3} \right),{\text{ }}\left( {5,{\text{ }}4} \right),{\text{ }}\left( {5,{\text{ }}5} \right),{\text{ }}\left( {5,{\text{ }}6} \right){\text{ }} = {\text{ }}6{\text{ }}possibilities.\]
\[\;\left( {6,{\text{ }}1} \right),{\text{ }}\left( {6,{\text{ }}2} \right),{\text{ }}\left( {6,{\text{ }}3} \right),{\text{ }}\left( {6,{\text{ }}4} \right),{\text{ }}\left( {6,{\text{ }}5} \right),{\text{ }}\left( {6,{\text{ }}6} \right){\text{ }} = {\text{ }}6{\text{ }}possibilities.\]
Total number of elements (possibilities) of set S are therefore,\[n\left( S \right) = 6 \times 6 = 36\]; i.e. six possibilities of second dice for each of the six possibilities of the first dice.
Note:
1) A sample space is usually denoted using set notation, and the possible ordered outcomes are listed as elements in the set.
2) The probability of an outcome E in a sample space S is a number P between 1 and 0 that measures the likelihood that E will occur on a single trial.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

