Question

Two digit are to be select at random from 1 to 9, if the sum is odd, what is the probability that one of the digit is 2

Answer 1

Total ways of selecting 2 numbers= 9C2
Sum is odd= 5C1×4C1=20
let a number is 2 then other one must be odd, so number of ways for other number = 5C1=5
P(E)=1/4

Answer 2

Two digits are to be selected at random from 1 to 9.
Total condition possiblity = {(1,2),(1,4),(1,6)(1,8),(2,1),(2,3),(2,5),(2,7)(2,9),(3,2),(3,4),(3,6),(3,8),(4,1),(4,3),(4,5),(4,7),(4,9),(5,2),(5,4),(5,6),(5,8),(6,1),(6,3),(6,5),(6,7),(6,9),(7,2),(7,4),(7,6),(7,8),(8,1),(8,3),(8,5),(8,7),(8,9),(9,2),(9,4),(9,6),(9,8)}.
= 40.

Favorable condition possiblity={(1,2),(2,1),(2,3),(2,5),(2,7),(2,9),(3,2),(5,2),(7,2),(9,2)}. = 10.

Therefore favorable condition probability =10/40.
=1/4.
Hope it helps.