A pair of number cubes is thrown. Find each probability given that their sum is greater than or equal to $9$.
- $P \left(\text{numbers match}\right)$
- $P \left(\text{numbers match or sum is even}\right)$
Number of elements in sample space $S = n \left(S\right) = 36$
Let $C$ be the event that the sum of the numbers on the two cubes is greater than or equal to $9$.
i.e. $\;$ $C = \left\{\left(3, \; 6\right), \; \left(4, \; 5\right), \; \left(4, \; 6\right), \; \left(5, \; 4\right), \; \left(5, \; 5\right), \right.$
$\hspace{3cm}$ $\left. \left(5, \; 6\right), \; \left(6, \; 3\right), \; \left(6, \; 4\right), \; \left(6, \; 5\right), \; \left(6, \; 6\right) \right\}$
$\therefore \;$ Number of elements in $C = n \left(C\right) = 10$
$\therefore \;$ $P \left(C\right) = \dfrac{n \left(C\right)}{n \left(S\right)} = \dfrac{10}{36}$
- Let $A$ be the event that the numbers on the two cubes match.
i.e. $\;$ $A = \left\{\left(1, \; 1\right), \; \left(2, \; 2\right), \; \left(3, \; 3\right), \; \left(4, \; 4\right), \; \left(5, \; 5\right), \; \left(6, \; 6\right) \right\}$
$\left(A \cap C\right) = \;$ event that the numbers on both the cubes match AND the sum of the numbers is greater than or equal to $9$
$\therefore \;$ $\left(A \cap C\right) = \left\{\left(5, \; 5\right), \; \left(6, \; 6\right) \right\}$
$\therefore \;$ Number of elements in $\left(A \cap C\right) = n \left(A \cap C\right) = 2$
$\therefore \;$ $P \left(A \cap C\right) = \dfrac{n \left(A \cap C\right)}{n \left(S\right)} = \dfrac{2}{36}$
$\therefore \;$ P(numbers on the two cubes match given that their sum is greater than or equal to 9)
$= P \left(A | C\right) = \dfrac{P \left(A \cap C\right)}{P \left(C\right)} = \dfrac{2/36}{10/36} = \dfrac{2}{10} = \dfrac{1}{5}$
- Let $B$ be the event that the numbers on the two cubes match or the sum is even
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. $
$\hspace{2cm}$ $\left. \left(1, \; 3\right), \; \left(1, \; 5\right), \; \left(2, \; 4\right), \; \left(2, \; 6\right), \; \left(3, \; 1\right), \; \left(3, \; 5\right) \right. $
$\hspace{3cm}$ $\left. \left(4, \; 2\right), \; \left(4, \; 6\right), \; \left(5, \; 1\right), \; \left(5, \; 3\right), \; \left(6, \; 2\right), \; \left(6, \; 4\right) \right\}$
$\left(B \cap C\right) = \;$ event that the numbers on the two cubes match or the sum is even AND the sum of the numbers is greater than or equal to $9$
$\therefore \;$ $\left(B \cap C\right) = \left\{\left(4, \; 6\right), \; \left(5, \; 5\right), \; \left(6, \; 4\right), \; \left(6, \; 6\right) \right\}$
$\therefore \;$ Number of elements in $\left(B \cap C\right) = n \left(B \cap C\right) = 4$
$\therefore \;$ $P \left(B \cap C\right) = \dfrac{n \left(B \cap C\right)}{n \left(S\right)} = \dfrac{4}{36}$
$\therefore \;$ P(numbers on the two cubes match or sum is even given that their sum is greater than or equal to 9)
$= P \left(B | C\right) = \dfrac{P \left(B \cap C\right)}{P \left(C\right)} = \dfrac{4/36}{10/36} = \dfrac{4}{10} = \dfrac{2}{5}$