
How do you determine the sample space for the experiment where a six sided die is tossed twice and the sum of the points is recorded?
Answer
541.8k+ views
Hint: The sample space of an experiment is defined as the set of all possible outcomes of that experiment. We will find all the possible combinations of numbers that can be on a six sided die when it is tossed twice. Then we will add the points to find the sum.
Complete step by step solution:
We know that a six sided die will possibly show the numbers $1,2,3,4,5$ and $6$ when it is tossed only once. This experiment $\left( 1 \right)$ has a sample space that contains only $6$ elements. The sample space of the experiment $\left( 1 \right)$ is \[{{\text{S}}_{1}}=\left\{ 1,2,3,4,5,6 \right\}.\]
Now, let us discuss the sample space of an experiment $\left( 2 \right)$ where a six sided die is tossed twice. Since the die contains the numbers $1,2,3,4,5$ and $6,$ the total number of the outcomes of the experiment $\left( 2 \right)$ is ${{6}^{2}}=36.$ That is, the cardinality of the sample space of the experiment $\left( 2 \right)$ is $36.$
Let us write the sample space of the experiment $\left( 2 \right).$
\[\begin{align}
& {{\text{S}}_{2}}=\left\{ \left( 1,1 \right),\left( 1,2 \right),\left( 1,3 \right),\left( 1,4 \right),\left( 1,5 \right),\left( 1,6 \right), \right. \\
& \left( 2,1 \right),\left( 2,2 \right),\left( 2,3 \right),\left( 2,4 \right),\left( 2,5 \right),\left( 2,6 \right) \\
& \left( 3,1 \right),\left( 3,2 \right),\left( 3,3 \right),\left( 3,4 \right),\left( 3,5 \right),\left( 3,6 \right), \\
& \left( 4,1 \right),\left( 4,2 \right),\left( 4,3 \right),\left( 4,4 \right),\left( 4,5 \right),\left( 4,6 \right), \\
& \left( 5,1 \right),\left( 5,2 \right),\left( 5,3 \right),\left( 5,4 \right),\left( 5,5 \right),\left( 5,6 \right), \\
& \left. \left( 6,1 \right),\left( 6,2 \right),\left( 6,3 \right),\left( 6,4 \right),\left( 6,5 \right),\left( 6,6 \right) \right\} \\
\end{align}\]
Now, the experiment $\left( 3 \right)$ in the question is asking us to find the sum of the points in each pair of elements in the sample space of the experiment $\left( 2 \right).$
By inspection, we can see that some of the numbers repeat. For example, $2+2=1+3=4.$ That is, there will be repeated numbers in the set that contains the sums of the points. So, we ignore the repeated terms and choose each of the terms only once.
Hence the sample space of the experiment where a six sided die is tossed twice and the sum of the points is recorded is $\text{S}=\left\{ 2,3,4,5,6,7,8,9,10,11,12 \right\}.$
Note: We will not write the same elements repeatedly in the set. If there are elements that repeat in a set, we can just ignore the repeated ones and choose all the elements only once. In the sample space ${{\text{S}}_{2}},$ the elements $\left( 1,2 \right)$ and $\left( 2,1 \right)$ are not the same. This is applicable for the rest of the elements also.
Complete step by step solution:
We know that a six sided die will possibly show the numbers $1,2,3,4,5$ and $6$ when it is tossed only once. This experiment $\left( 1 \right)$ has a sample space that contains only $6$ elements. The sample space of the experiment $\left( 1 \right)$ is \[{{\text{S}}_{1}}=\left\{ 1,2,3,4,5,6 \right\}.\]
Now, let us discuss the sample space of an experiment $\left( 2 \right)$ where a six sided die is tossed twice. Since the die contains the numbers $1,2,3,4,5$ and $6,$ the total number of the outcomes of the experiment $\left( 2 \right)$ is ${{6}^{2}}=36.$ That is, the cardinality of the sample space of the experiment $\left( 2 \right)$ is $36.$
Let us write the sample space of the experiment $\left( 2 \right).$
\[\begin{align}
& {{\text{S}}_{2}}=\left\{ \left( 1,1 \right),\left( 1,2 \right),\left( 1,3 \right),\left( 1,4 \right),\left( 1,5 \right),\left( 1,6 \right), \right. \\
& \left( 2,1 \right),\left( 2,2 \right),\left( 2,3 \right),\left( 2,4 \right),\left( 2,5 \right),\left( 2,6 \right) \\
& \left( 3,1 \right),\left( 3,2 \right),\left( 3,3 \right),\left( 3,4 \right),\left( 3,5 \right),\left( 3,6 \right), \\
& \left( 4,1 \right),\left( 4,2 \right),\left( 4,3 \right),\left( 4,4 \right),\left( 4,5 \right),\left( 4,6 \right), \\
& \left( 5,1 \right),\left( 5,2 \right),\left( 5,3 \right),\left( 5,4 \right),\left( 5,5 \right),\left( 5,6 \right), \\
& \left. \left( 6,1 \right),\left( 6,2 \right),\left( 6,3 \right),\left( 6,4 \right),\left( 6,5 \right),\left( 6,6 \right) \right\} \\
\end{align}\]
Now, the experiment $\left( 3 \right)$ in the question is asking us to find the sum of the points in each pair of elements in the sample space of the experiment $\left( 2 \right).$
By inspection, we can see that some of the numbers repeat. For example, $2+2=1+3=4.$ That is, there will be repeated numbers in the set that contains the sums of the points. So, we ignore the repeated terms and choose each of the terms only once.
Hence the sample space of the experiment where a six sided die is tossed twice and the sum of the points is recorded is $\text{S}=\left\{ 2,3,4,5,6,7,8,9,10,11,12 \right\}.$
Note: We will not write the same elements repeatedly in the set. If there are elements that repeat in a set, we can just ignore the repeated ones and choose all the elements only once. In the sample space ${{\text{S}}_{2}},$ the elements $\left( 1,2 \right)$ and $\left( 2,1 \right)$ are not the same. This is applicable for the rest of the elements also.
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