Two game tiles, numbered $1$ through $9$, are selected at random from a box without replacement. If their sum is even, what is the probability that both the numbers are odd?
$2$ game tiles can be selected from $9$ tiles in ${^{9}}{P}_{2} = \dfrac{9!}{7!} = 9 \times 8 = 72 \;$ ways
$\therefore \;$ Number of elements in sample space $S = n \left(S\right) = 72$
Let $A$ be the event that both the tiles are odd.
Let $B$ be the event that the sum of numbers on the two tiles is even.
This happens when both the selected tiles are odd \textbf{OR} both are even.
There are $5$ odd tiles (1, 3, 5, 7, 9) and $4$ even tiles (2,4,6,8).
$2$ Odd tiles can be selected from $5$ tiles in ${^{5}}{P}_{2} = \dfrac{5!}{3!} = 5 \times 4 = 20 \;$ ways
$2$ even tiles can be selected from $4$ tiles in ${^{4}}{P}_{2} = \dfrac{4!}{2!} = 4 \times 3 = 12 \;$ ways
$\therefore \;$ Number of elements in $B = n \left(B\right) = 20 + 12 = 32$
$\therefore \;$ $P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{32}{72}$
$\left(A \cap B\right)$ is the event that both the tiles are odd numbered \textbf{AND} their sum is even.
$\therefore \;$ $n \left(A \cap B\right) = 20$
$\therefore \;$ $P \left(A \cap B\right) = \dfrac{n \left(A \cap B\right)}{n \left(S\right)} = \dfrac{20}{72}$
Now, P(that both the numbers are odd given that their sum is even)
$= P \left(A | B\right) = \dfrac{P \left(A \cap B\right)}{P \left(B\right)} = \dfrac{20/72}{32/72} = \dfrac{20}{32} = \dfrac{5}{8}$