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Hint: To find the sample space in the above question, we will find the sample space of two events separately i.e. we will find outcomes of both the events separately. Then we will find the total number of outcomes according to the new sample space formed from the combination of these events.
Complete step-by-step answer:
Before solving this question, we must first know what is a sample space. A sample space of an experiment or a random trial is the set of all possible outcomes or results of that experiment. A sample space is usually denoted using set notation (S), and the possible ordered outcomes are listed as elements in the set. Now in the question, we are given two events, one of them is the tossing of a coin. When a coin is tossed, we get two outcomes: either a heads will come or tails will come. Let \({S_1}\) be the sample space of tossing of a coin. Thus, we have:
\({S_1}{\rm{ = }}\left\{ {Heads,{\rm{ Tails}}} \right\}\)
Another event given in question is the rolling of dice. When a dice is rolled, we get six outcomes: 1, 2, 3, 4, 5 and 6. Let \({S_2}\) be the sample space of rolling of dice. Thus, we have:
\({S_2}{\rm{ = }}\left\{ {1,2,3,4,5,6} \right\}\)
Now, we have to combine these events and now both the events happen simultaneously. Let us suppose that heads comes in \({S_1}\) then \({S_2}\) can have 1, 2, 3, 4, 5 or 6. Similarly, if tails comes in \({S_1}\) then \({S_2}\) can have 1, 2, 3, 4, 5 or 6. Thus the sample space when these events happen simultaneously is denoted by S. Thus we have:
\(S{\rm{ = }}\left\{ \begin{array}{l}\left( {Heads,{\rm{1}}} \right),\left( {Heads,2} \right),\left( {Heads,3} \right),\left( {Heads,4} \right),\left( {Heads,5} \right),\left( {Heads,6} \right),\\\left( {Tails,1} \right),\left( {Tails,2} \right),\left( {Tails,3} \right),\left( {Tails,4} \right),\left( {Tails,5} \right),\left( {Tails,6} \right)\end{array} \right\}\)
Now, we have calculated n(S) i.e. the total number of outcomes in S. Thus, on counting we get to know that, \[{\rm{n}}\left( {\rm{S}} \right) = {\rm{12}}\].
Note: We can apply this method of combining two sample spaces only when both the events are independent events i.e. the happening of one event (rolling of dice) does not affect the happening of another (tossing of coin).
Complete step-by-step answer:
Before solving this question, we must first know what is a sample space. A sample space of an experiment or a random trial is the set of all possible outcomes or results of that experiment. A sample space is usually denoted using set notation (S), and the possible ordered outcomes are listed as elements in the set. Now in the question, we are given two events, one of them is the tossing of a coin. When a coin is tossed, we get two outcomes: either a heads will come or tails will come. Let \({S_1}\) be the sample space of tossing of a coin. Thus, we have:
\({S_1}{\rm{ = }}\left\{ {Heads,{\rm{ Tails}}} \right\}\)
Another event given in question is the rolling of dice. When a dice is rolled, we get six outcomes: 1, 2, 3, 4, 5 and 6. Let \({S_2}\) be the sample space of rolling of dice. Thus, we have:
\({S_2}{\rm{ = }}\left\{ {1,2,3,4,5,6} \right\}\)
Now, we have to combine these events and now both the events happen simultaneously. Let us suppose that heads comes in \({S_1}\) then \({S_2}\) can have 1, 2, 3, 4, 5 or 6. Similarly, if tails comes in \({S_1}\) then \({S_2}\) can have 1, 2, 3, 4, 5 or 6. Thus the sample space when these events happen simultaneously is denoted by S. Thus we have:
\(S{\rm{ = }}\left\{ \begin{array}{l}\left( {Heads,{\rm{1}}} \right),\left( {Heads,2} \right),\left( {Heads,3} \right),\left( {Heads,4} \right),\left( {Heads,5} \right),\left( {Heads,6} \right),\\\left( {Tails,1} \right),\left( {Tails,2} \right),\left( {Tails,3} \right),\left( {Tails,4} \right),\left( {Tails,5} \right),\left( {Tails,6} \right)\end{array} \right\}\)
Now, we have calculated n(S) i.e. the total number of outcomes in S. Thus, on counting we get to know that, \[{\rm{n}}\left( {\rm{S}} \right) = {\rm{12}}\].
Note: We can apply this method of combining two sample spaces only when both the events are independent events i.e. the happening of one event (rolling of dice) does not affect the happening of another (tossing of coin).
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