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Hint: Now in the question it consists of two events one after the other and we have to represent it in the sample space in such a way that each element represents these two simultaneously occurring events and for that we will use ordered pairs concept. Event 1 is the selecting of one die and the sample space is \[{{S}_{1}}=\left\{ Red,White,Blue \right\}\], and the Event 2 is the noting down the number on the uppermost face of the die hence the sample space is \[{{S}_{2}}=\left\{ 1,2,3,4,5,6 \right\}\]. Now, we can combine these sample spaces to get our final sample space.

Sample space is the set of all possible outcomes of an experiment and is used to calculate the probability of an event related to the respective experiment. The elements of sample space represent the possible result of each trial. The number of favourable outcomes are chosen from this set along with the total possible outcomes.

Now as we want to find the sample space of the experiment which contains two simultaneously occurring events one of which is selecting a die at random from the bag and the other following it is throwing the die which came out and note down the number on the uppermost face.

Now we are going to construct the sample space for the Event 1 which is selecting a die from the bag,

Let us consider the sample space of Event 1 be represented by \[{{S}_{1}}\],

\[{{S}_{1}}=\left\{ Red,White,Blue \right\}\]

Let us translate the colours for easiness and hence red be R, white be W and blue be B, so now the sample space is,

\[{{S}_{1}}=\left\{ R,W,B \right\}\]

As the probability of selecting each die is the same.

Now the sample space for Event 2 which is to note down the number appearing on uppermost face of the die is,

Let the sample space of Event 2 be represented by \[{{S}_{2}}\],

\[{{S}_{2}}=\left\{ 1,2,3,4,5,6 \right\}\]

As a die has six outcomes and each having equal probability of occurring on the uppermost face of the die.

Now to merge the two events and get the two events outcomes represented in such a way that every element of the final sample space represents the outcome of both the events and at once.

For this to happen we go for the ordered pair concept which is that we represent each ordered pair has a particular order and it cannot be changed but if changed then the interchanged and original have different meanings.

General way to represent an ordered pair is \[\left( x,y \right)\] where \[x\] represents the Event 1 outcomes and \[y\] represents the Event 2 outcomes. Event 1 occurs prior to Event 2 and they are represented in the same way in ordered pairs.

So the values of \[x\] are the values of set \[{{S}_{1}}\], and the values of \[y\] are the values of set \[{{S}_{2}}\].

So,

\[\begin{align}

& x\in {{S}_{1}}=\left\{ R,W,B \right\} \\

& y\in {{S}_{2}}=\left\{ 1,2,3,4,5,6 \right\} \\

\end{align}\]

Hence, let the final sample space be \[{{S}_{f}}\],

\[{{S}_{f}}=\left\{ \left( R,1 \right),\left( R,2 \right),\left( R,3 \right),\left( R,4 \right),\left( R,5 \right),\left( R,6 \right),\left( W,1 \right),\left( W,2 \right),\left( W,3 \right),\left( W,4 \right),\left( W,5 \right),\left( W,6 \right),\left( B,1 \right),\left( B,2 \right),\left( B,3 \right),\left( B,4 \right),\left( B,5 \right),\left( B,6 \right) \right\}\]

\[{{S}_{f}}=\left\{ \begin{align}

& \left( R,1 \right),\left( R,2 \right),\left( R,3 \right),\left( R,4 \right),\left( R,5 \right),\left( R,6 \right) \\

& ,\left( W,1 \right),\left( W,2 \right),\left( W,3 \right),\left( W,4 \right),\left( W,5 \right),\left( W,6 \right) \\

& ,\left( B,1 \right),\left( B,2 \right),\left( B,3 \right),\left( B,4 \right),\left( B,5 \right),\left( B,6 \right) \\

\end{align} \right\}\]

As when we take out a die at random we can get any number from 1 to 6 when we throw it and hence to represent from which colour die it came we have to write die color also as that is also an event.

For every die colour there are six outcomes so six ordered pairs are related to each die of the same colour.

Note: An additional concept to remember is that number of ordered pairs is equal to the product of number of elements of both the sets like in this case above we have \[n\left( {{S}_{1}} \right)=3\] and \[n\left( {{S}_{2}} \right)=6\], so according to this concept \[n\left( {{S}_{f}} \right)=n\left( {{S}_{1}} \right)\times n\left( {{S}_{2}} \right)=3\times 6=18\] which is same as we have calculated in the set \[{{S}_{f}}\]. You can also use it to verify the number of terms in your answer and what it should be according to the formula.

It will be a mistake if you interchange the \[x\] and \[y\] positions in the ordered pair because it is a rule that \[\left( x,y \right)\ne \left( y,x \right)\] as the position also has a significance which shows position of \[x\] is Domain and position of \[y\] is Co-Domain or Range of a function.

Whenever there are two events in an experiment which occur one after the other always use ordered pairs to solve these questions. Let us take an example like if two die are thrown simultaneously then the output will be 36 elements and each element is an ordered pair of numbers. Because both sets have six elements each and the ordered pairs containing the set contains the number of elements equal to that of the product of the number of elements of the sets mentioned above.

__Complete step-by-step answer:__Sample space is the set of all possible outcomes of an experiment and is used to calculate the probability of an event related to the respective experiment. The elements of sample space represent the possible result of each trial. The number of favourable outcomes are chosen from this set along with the total possible outcomes.

Now as we want to find the sample space of the experiment which contains two simultaneously occurring events one of which is selecting a die at random from the bag and the other following it is throwing the die which came out and note down the number on the uppermost face.

Now we are going to construct the sample space for the Event 1 which is selecting a die from the bag,

Let us consider the sample space of Event 1 be represented by \[{{S}_{1}}\],

\[{{S}_{1}}=\left\{ Red,White,Blue \right\}\]

Let us translate the colours for easiness and hence red be R, white be W and blue be B, so now the sample space is,

\[{{S}_{1}}=\left\{ R,W,B \right\}\]

As the probability of selecting each die is the same.

Now the sample space for Event 2 which is to note down the number appearing on uppermost face of the die is,

Let the sample space of Event 2 be represented by \[{{S}_{2}}\],

\[{{S}_{2}}=\left\{ 1,2,3,4,5,6 \right\}\]

As a die has six outcomes and each having equal probability of occurring on the uppermost face of the die.

Now to merge the two events and get the two events outcomes represented in such a way that every element of the final sample space represents the outcome of both the events and at once.

For this to happen we go for the ordered pair concept which is that we represent each ordered pair has a particular order and it cannot be changed but if changed then the interchanged and original have different meanings.

General way to represent an ordered pair is \[\left( x,y \right)\] where \[x\] represents the Event 1 outcomes and \[y\] represents the Event 2 outcomes. Event 1 occurs prior to Event 2 and they are represented in the same way in ordered pairs.

So the values of \[x\] are the values of set \[{{S}_{1}}\], and the values of \[y\] are the values of set \[{{S}_{2}}\].

So,

\[\begin{align}

& x\in {{S}_{1}}=\left\{ R,W,B \right\} \\

& y\in {{S}_{2}}=\left\{ 1,2,3,4,5,6 \right\} \\

\end{align}\]

Hence, let the final sample space be \[{{S}_{f}}\],

\[{{S}_{f}}=\left\{ \left( R,1 \right),\left( R,2 \right),\left( R,3 \right),\left( R,4 \right),\left( R,5 \right),\left( R,6 \right),\left( W,1 \right),\left( W,2 \right),\left( W,3 \right),\left( W,4 \right),\left( W,5 \right),\left( W,6 \right),\left( B,1 \right),\left( B,2 \right),\left( B,3 \right),\left( B,4 \right),\left( B,5 \right),\left( B,6 \right) \right\}\]

\[{{S}_{f}}=\left\{ \begin{align}

& \left( R,1 \right),\left( R,2 \right),\left( R,3 \right),\left( R,4 \right),\left( R,5 \right),\left( R,6 \right) \\

& ,\left( W,1 \right),\left( W,2 \right),\left( W,3 \right),\left( W,4 \right),\left( W,5 \right),\left( W,6 \right) \\

& ,\left( B,1 \right),\left( B,2 \right),\left( B,3 \right),\left( B,4 \right),\left( B,5 \right),\left( B,6 \right) \\

\end{align} \right\}\]

As when we take out a die at random we can get any number from 1 to 6 when we throw it and hence to represent from which colour die it came we have to write die color also as that is also an event.

For every die colour there are six outcomes so six ordered pairs are related to each die of the same colour.

Note: An additional concept to remember is that number of ordered pairs is equal to the product of number of elements of both the sets like in this case above we have \[n\left( {{S}_{1}} \right)=3\] and \[n\left( {{S}_{2}} \right)=6\], so according to this concept \[n\left( {{S}_{f}} \right)=n\left( {{S}_{1}} \right)\times n\left( {{S}_{2}} \right)=3\times 6=18\] which is same as we have calculated in the set \[{{S}_{f}}\]. You can also use it to verify the number of terms in your answer and what it should be according to the formula.

It will be a mistake if you interchange the \[x\] and \[y\] positions in the ordered pair because it is a rule that \[\left( x,y \right)\ne \left( y,x \right)\] as the position also has a significance which shows position of \[x\] is Domain and position of \[y\] is Co-Domain or Range of a function.

Whenever there are two events in an experiment which occur one after the other always use ordered pairs to solve these questions. Let us take an example like if two die are thrown simultaneously then the output will be 36 elements and each element is an ordered pair of numbers. Because both sets have six elements each and the ordered pairs containing the set contains the number of elements equal to that of the product of the number of elements of the sets mentioned above.

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