Question

Two dice are thrown at the same time. Complete the following table.

Event: Sum on \[2\] dice	Probability
\[2\]	\[\dfrac{1}{{36}}\]
\[3\]
\[4\]
\[5\]
\[6\]
\[7\]
\[8\]	\[\dfrac{5}{{36}}\]
\[9\]
\[10\]
\[11\]
\[12\]	\[\dfrac{1}{{36}}\]

Answer 1

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

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.