Two digits selected from $1$ to $9$ and the sum is odd , what is the probability that one of the digits is $2$ The answer given in the book is $1/4$ but i don't understand how that can be so. What i understand from the question is this based on conditional probability. i have to find :
P(one number being $2$ , GIVEN sum of two numbers is odd) = P(2 $\cap$ odd sum)$\div$ P(odd sum)
but i'm not sure how to do/calculate this. Any help? Also , do correct me if my understanding is wrong.
edit : Repeats are not allowed in selection. Forgot to add this.
 A: Since the sum is odd, one number is odd and the other is even. Thus there are $2\times 5\times 4 = 40$ choices for selecting a pair of number whose sum is odd. Of these pairs, there are $2\times 5 = 10$ pairs with one of the numbers is $2$. Thus the answer is $\dfrac{10}{40} = \dfrac{1}{4}$.
A: My suggestion don't go to formulas try to work it out
Consider the ordered pairs $(1,2),(1,4),(1,6),(1,8)$ when $1^{st}$ element of ordered pair is taken as$1$
Consider the ordered pairs $(2,1),(2,3),(2,5),(2,7),(2,9)$ when $1^{st}$ element of ordered pair is taken as$2$ 
going on in this way see that when we have the $1^{st}$ element of the ordered pair odd we get $4$ elements  and when even we get $5$ elements.Since from $1$ to $9$ we have  $5$ odd and $4$ even elements  we get number of elemnts to be $5\times +4\times 5=40 $  elements.total number of possible outcomes =40
No. of elements having an element 2 is 5+5=10 since each when the 1st lement of the ordered pair is odd we can have 2as the other element in the pair otherwise not.And another 5 as I have explicitly written down viz.  $(2,1),(2,3),(2,5),(2,7),(2,9)$
Favourable oucomes =10
So prob.=$10/40=1/4$
