Question

In a throw of a pair of dice, the probability of getting a doublet is:
A) \[\dfrac{1}{2}\]
B) \[\dfrac{1}{3}\]
C) \[\dfrac{1}{6}\]
D) \[\dfrac{5}{6}\]

Answer 1

Hint:
Here, we have to find the probability of getting a doublet, in a throw of a pair of dice. First, we will find the sample space, then we will find the number of favorable outcomes. We will then substitute the values in the formula to find the required probability. Probability means the possibility of an event to happen.

Formula Used:
The Probability is given by the formula \[P\left( E \right){\rm{ = }}\dfrac{{n\left( E \right)}}{{n\left( S \right)}}\], where \[n\left( E \right)\] is the number of favorable outcome and \[n\left( S \right)\] is the total number of outcomes.

Complete step by step solution:
We will first determine the sample space for a pair of dice.
\[S = \left\{ {\begin{array}{*{20}{c}}{\left( {1,1} \right)}&{\left( {1,2} \right)}&{\left( {1,3} \right)}&{\left( {1,4} \right)}&{\left( {1,5} \right)}&{\left( {1,6} \right)}\\{\left( {2,1} \right)}&{\left( {2,2} \right)}&{\left( {2,3} \right)}&{\left( {2,4} \right)}&{\left( {2,5} \right)}&{\left( {2,6} \right)}\\{\left( {3,1} \right)}&{\left( {3,2} \right)}&{\left( {3,3} \right)}&{\left( {3,4} \right)}&{\left( {3,5} \right)}&{\left( {3,6} \right)}\\{\left( {4,1} \right)}&{\left( {4,2} \right)}&{\left( {4,3} \right)}&{\left( {4,4} \right)}&{\left( {4,5} \right)}&{\left( {4,6} \right)}\\{\left( {5,1} \right)}&{\left( {5,2} \right)}&{\left( {5,3} \right)}&{\left( {5,4} \right)}&{\left( {5,5} \right)}&{\left( {5,6} \right)}\\{\left( {6,1} \right)}&{\left( {6,2} \right)}&{\left( {6,3} \right)}&{\left( {6,4} \right)}&{\left( {6,5} \right)}&{\left( {6,6} \right)}\end{array}} \right\}\]
Total Number of Outcomes, \[n\left( S \right) = 36\]
Let \[E\] be the event of getting a doublet thrown in a pair of dice. Therefore,
\[E = \left\{ {\left( {1,1} \right),\left( {2,2} \right),\left( {3,3} \right),\left( {4,4} \right),\left( {5,5} \right),\left( {6,6} \right)} \right\}\]
Number of Favorable Outcomes, \[n\left( E \right) = 6\]
Substituting the values of \[n\left( E \right) = 6\] and \[n\left( S \right) = 36\] in the formula \[P\left( E \right){\rm{ = }}\dfrac{{n\left( E \right)}}{{n\left( S \right)}}\], we get
\[P\left( E \right) = \dfrac{6}{{36}}\]
Dividing both numerator and denominator by 6, we get
\[ \Rightarrow P\left( E \right) = \dfrac{1}{6}\]
Therefore, in a throw of a pair of dice, the probability of getting a doublet is \[\dfrac{1}{6}\].

Hence, option C is the correct answer.

Additional Information:
Properties of Probability include that the probability should be positive or zero. The sum of all the probabilities of possible outcomes is 1. An event which happens for sure is called a sure event and the probability of a sure event should be 1. An event which will not occur ever is called an impossible event and the probability of an impossible event should be 0.

Note:
We can check the probability of an event since the probability always lies between 0 and 1. It can never be negative. The sum of the probabilities of an event and its complementary event should be one. So, by using this property, we get
\[P\left( E \right) + P\left( {E'} \right) = 1\] ………………………..\[\left( 1 \right)\]
Let \[E'\] be the event of not getting a doublet. So,
\[n\left( {E'} \right) = 30\]
So, the probability of not getting a doublet will be:
\[P\left( {E'} \right) = \dfrac{{n\left( {E'} \right)}}{{n\left( S \right)}} = \dfrac{{30}}{{36}}\]
Rewriting the equation \[\left( 1 \right)\], we get
\[ \Rightarrow P\left( E \right) = 1 - P\left( {E'} \right)\]
Substituting the value of \[P\left( {E'} \right)\] in the above equation, we get
\[ \Rightarrow P\left( E \right) = 1 - \dfrac{{30}}{{36}}\]
By cross-multiplication, we get
\[ \Rightarrow P\left( E \right) = \dfrac{{36 - 30}}{{36}}\]
\[ \Rightarrow P\left( E \right) = \dfrac{6}{{36}} = \dfrac{1}{6}\]