Two dice are thrown at the same time. Complete the following table.
Event: Sum on \[2\] dice Probability \[2\] \[\dfrac{1}{{36}}\] \[3\] \[4\] \[5\] \[6\] \[7\] \[8\] \[\dfrac{5}{{36}}\] \[9\] \[10\] \[11\] \[12\] \[\dfrac{1}{{36}}\]
| Event: Sum on \[2\] dice | Probability |
| \[2\] | \[\dfrac{1}{{36}}\] |
| \[3\] | |
| \[4\] | |
| \[5\] | |
| \[6\] | |
| \[7\] | |
| \[8\] | \[\dfrac{5}{{36}}\] |
| \[9\] | |
| \[10\] | |
| \[11\] | |
| \[12\] | \[\dfrac{1}{{36}}\] |
Answer
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Hint: Here, we have to complete the table by using the concept of probability. Probability means possibility. Probability is a measure of the likelihood of an event to occur. An event is an outcome or defined collection of outcomes of a random experiment.
Formula used:
We will use the formula for probability \[P(E) = \dfrac{{n(E)}}{{n(S)}}\], where \[n\left( E \right)\] is the number of events and \[n\left( S \right)\] is the total number of sample space.
Complete step-by-step answer:
We have to determine the sample space of two dice
\[S = \left\{ {\begin{array}{*{20}{c}}{\left( {1,1} \right)}&{\left( {1,2} \right)}&{\left( {1,3} \right)}&{\left( {1,4} \right)}&{\left( {1,5} \right)}&{\left( {1,6} \right)}\\{\left( {2,1} \right)}&{\left( {2,2} \right)}&{\left( {2,3} \right)}&{\left( {2,4} \right)}&{\left( {2,5} \right)}&{\left( {2,6} \right)}\\{\left( {3,1} \right)}&{\left( {3,2} \right)}&{\left( {3,3} \right)}&{\left( {3,4} \right)}&{\left( {3,5} \right)}&{\left( {3,6} \right)}\\{\left( {4,1} \right)}&{\left( {4,2} \right)}&{\left( {4,3} \right)}&{\left( {4,4} \right)}&{\left( {4,5} \right)}&{\left( {4,6} \right)}\\{\left( {5,1} \right)}&{\left( {5,2} \right)}&{\left( {5,3} \right)}&{\left( {5,4} \right)}&{\left( {5,5} \right)}&{\left( {5,6} \right)}\\{\left( {6,1} \right)}&{\left( {6,2} \right)}&{\left( {6,3} \right)}&{\left( {6,4} \right)}&{\left( {6,5} \right)}&{\left( {6,6} \right)}\end{array}} \right\}\]
Therefore, we got number of samples as:
\[n(S) = 36\]
Let \[E\] be the event of getting the sum equal to \[3\].
\[ \Rightarrow n(E) = \left( {1,2} \right),\left( {2,1} \right) = 2\]
By using the formula \[P(E) = \dfrac{{n(E)}}{{n(S)}}\], we have
\[ \Rightarrow P(E) = \dfrac{2}{{36}}\]
Let \[E\] be the event of getting the sum equal to 4.
\[ \Rightarrow n(E) = \left( {1,3} \right),\left( {2,2} \right),\left( {3,1} \right) = 3\]
By using the formula \[P(E) = \dfrac{{n(E)}}{{n(S)}}\], we have
\[ \Rightarrow P(E) = \dfrac{3}{{36}}\]
Let \[E\] be the event of getting the sum equal to 5.
\[ \Rightarrow n(E) = \left( {1,4} \right)\left( {2,3} \right)\left( {3,2} \right)\left( {4,1} \right) = 4\]
By using the formula \[P(E) = \dfrac{{n(E)}}{{n(S)}}\], we have
\[ \Rightarrow P(E) = \dfrac{4}{{36}}\]
Let \[E\] be the event of getting the sum equal to 6.
\[ \Rightarrow n(E) = \left( {1,5} \right),\left( {2,4} \right),\left( {3,3} \right),\left( {4,2} \right),\left( {5,1} \right) = 5\]
By using the probability formula, we have
\[ \Rightarrow P(E) = \dfrac{5}{{36}}\]
Let \[E\] be the event of getting the sum equal to 7.
\[ \Rightarrow n(E) = \left( {1,6} \right),\left( {2,5} \right),\left( {3,4} \right),\left( {4,3} \right),\left( {5,2} \right),\left( {6,1} \right) = 6\]
By using the formula, we have
\[ \Rightarrow P(E) = \dfrac{6}{{36}}\]
Let \[E\] be the event of getting the sum equal to 9.
\[ \Rightarrow n(E) = \left( {3,6} \right),\left( {4,5} \right),\left( {5,4} \right),\left( {6,3} \right) = 4\]
By using the probability formula, we have
\[ \Rightarrow P(E) = \dfrac{4}{{36}}\]
Let \[E\] be the event of getting the sum equal to 10.
\[ \Rightarrow n(E) = \left( {4,6} \right),\left( {5,5} \right),\left( {6,4} \right) = 3\]
By using the probability formula, we have
\[ \Rightarrow P(E) = \dfrac{3}{{36}}\]
Let \[E\] be the event of getting the sum equal to 11.
\[ \Rightarrow n(E) = \left( {5,6} \right),\left( {6,5} \right) = 2\]
By using the probability formula, we have
\[ \Rightarrow P(E) = \dfrac{2}{{36}}\]
Therefore, we can complete the table as
Note: We know that the dice has six faces with numbers from 1 to 6.So, the sample space for the dice related problems can be found using the formula \[{6^n}\]where \[n\] denotes the number of dices thrown at a time simultaneously. The probability of all events in a sample space adds up to 1. Sample space is the set of all the possible outcomes to occur in any trial. Outcome is the possible result of an experiment.
Formula used:
We will use the formula for probability \[P(E) = \dfrac{{n(E)}}{{n(S)}}\], where \[n\left( E \right)\] is the number of events and \[n\left( S \right)\] is the total number of sample space.
Complete step-by-step answer:
We have to determine the sample space of two dice
\[S = \left\{ {\begin{array}{*{20}{c}}{\left( {1,1} \right)}&{\left( {1,2} \right)}&{\left( {1,3} \right)}&{\left( {1,4} \right)}&{\left( {1,5} \right)}&{\left( {1,6} \right)}\\{\left( {2,1} \right)}&{\left( {2,2} \right)}&{\left( {2,3} \right)}&{\left( {2,4} \right)}&{\left( {2,5} \right)}&{\left( {2,6} \right)}\\{\left( {3,1} \right)}&{\left( {3,2} \right)}&{\left( {3,3} \right)}&{\left( {3,4} \right)}&{\left( {3,5} \right)}&{\left( {3,6} \right)}\\{\left( {4,1} \right)}&{\left( {4,2} \right)}&{\left( {4,3} \right)}&{\left( {4,4} \right)}&{\left( {4,5} \right)}&{\left( {4,6} \right)}\\{\left( {5,1} \right)}&{\left( {5,2} \right)}&{\left( {5,3} \right)}&{\left( {5,4} \right)}&{\left( {5,5} \right)}&{\left( {5,6} \right)}\\{\left( {6,1} \right)}&{\left( {6,2} \right)}&{\left( {6,3} \right)}&{\left( {6,4} \right)}&{\left( {6,5} \right)}&{\left( {6,6} \right)}\end{array}} \right\}\]
Therefore, we got number of samples as:
\[n(S) = 36\]
Let \[E\] be the event of getting the sum equal to \[3\].
\[ \Rightarrow n(E) = \left( {1,2} \right),\left( {2,1} \right) = 2\]
By using the formula \[P(E) = \dfrac{{n(E)}}{{n(S)}}\], we have
\[ \Rightarrow P(E) = \dfrac{2}{{36}}\]
Let \[E\] be the event of getting the sum equal to 4.
\[ \Rightarrow n(E) = \left( {1,3} \right),\left( {2,2} \right),\left( {3,1} \right) = 3\]
By using the formula \[P(E) = \dfrac{{n(E)}}{{n(S)}}\], we have
\[ \Rightarrow P(E) = \dfrac{3}{{36}}\]
Let \[E\] be the event of getting the sum equal to 5.
\[ \Rightarrow n(E) = \left( {1,4} \right)\left( {2,3} \right)\left( {3,2} \right)\left( {4,1} \right) = 4\]
By using the formula \[P(E) = \dfrac{{n(E)}}{{n(S)}}\], we have
\[ \Rightarrow P(E) = \dfrac{4}{{36}}\]
Let \[E\] be the event of getting the sum equal to 6.
\[ \Rightarrow n(E) = \left( {1,5} \right),\left( {2,4} \right),\left( {3,3} \right),\left( {4,2} \right),\left( {5,1} \right) = 5\]
By using the probability formula, we have
\[ \Rightarrow P(E) = \dfrac{5}{{36}}\]
Let \[E\] be the event of getting the sum equal to 7.
\[ \Rightarrow n(E) = \left( {1,6} \right),\left( {2,5} \right),\left( {3,4} \right),\left( {4,3} \right),\left( {5,2} \right),\left( {6,1} \right) = 6\]
By using the formula, we have
\[ \Rightarrow P(E) = \dfrac{6}{{36}}\]
Let \[E\] be the event of getting the sum equal to 9.
\[ \Rightarrow n(E) = \left( {3,6} \right),\left( {4,5} \right),\left( {5,4} \right),\left( {6,3} \right) = 4\]
By using the probability formula, we have
\[ \Rightarrow P(E) = \dfrac{4}{{36}}\]
Let \[E\] be the event of getting the sum equal to 10.
\[ \Rightarrow n(E) = \left( {4,6} \right),\left( {5,5} \right),\left( {6,4} \right) = 3\]
By using the probability formula, we have
\[ \Rightarrow P(E) = \dfrac{3}{{36}}\]
Let \[E\] be the event of getting the sum equal to 11.
\[ \Rightarrow n(E) = \left( {5,6} \right),\left( {6,5} \right) = 2\]
By using the probability formula, we have
\[ \Rightarrow P(E) = \dfrac{2}{{36}}\]
Therefore, we can complete the table as
| Event: Sum on \[2\] dice | Probability |
| \[2\] | \[\dfrac{1}{{36}}\] |
| \[3\] | \[\dfrac{2}{{36}}\] |
| \[4\] | \[\dfrac{3}{{36}}\] |
| \[5\] | \[\dfrac{4}{{36}}\] |
| \[6\] | \[\dfrac{5}{{36}}\] |
| \[7\] | \[\dfrac{6}{{36}}\] |
| \[8\] | \[\dfrac{5}{{36}}\] |
| \[9\] | \[\dfrac{4}{{36}}\] |
| \[10\] | \[\dfrac{3}{{36}}\] |
| \[11\] | \[\dfrac{2}{{36}}\] |
| \[12\] | \[\dfrac{1}{{36}}\] |
Note: We know that the dice has six faces with numbers from 1 to 6.So, the sample space for the dice related problems can be found using the formula \[{6^n}\]where \[n\] denotes the number of dices thrown at a time simultaneously. The probability of all events in a sample space adds up to 1. Sample space is the set of all the possible outcomes to occur in any trial. Outcome is the possible result of an experiment.
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