Probability

In a single throw of two dice, what is the probability of getting

1. a total of $9$;
2. a doublet;
3. $5$ on one die and $6$ on the other.

Let $S =$ event of throwing two dice.

Number of elements in sample space $= n \left(S\right) = 36$

1. Let $A =$ event of getting a total of $9$ when two dice are thrown
   
   i.e. $\;$ $A = \left\{\left(3,6\right), \left(4,5\right), \left(5,4\right), \left(6,3\right) \right\}$
   
   $\therefore \;$ Number of elements in $A = n \left(A\right) = 4$
   
   $\therefore \;$ Probability of event $A = P \left(A\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{4}{36} = \dfrac{1}{9}$



2. Let $B =$ event of getting a doublet when two dice are thrown
   
   i.e. $\;$ $B = \left\{\left(1,1\right), \left(2,2\right), \left(3,3\right), \left(4,4\right), \left(5,5\right), \left(6,6\right) \right\}$
   
   $\therefore \;$ Number of elements in $B = n \left(B\right) = 6$
   
   $\therefore \;$ Probability of event $B = P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{6}{36} = \dfrac{1}{6}$



3. Let $C =$ event of getting a $5$ on one die and $6$ on the other
   
   i.e. $\;$ $C = \left\{\left(5,6\right), \left(6,5\right) \right\}$
   
   $\therefore \;$ Number of elements in $C = n \left(C\right) = 2$
   
   $\therefore \;$ Probability of event $C = P \left(C\right) = \dfrac{n \left(C\right)}{n \left(S\right)} = \dfrac{2}{36} = \dfrac{1}{18}$