I would think you would just do:
The end results are the same but the odd results differ.
A jar contains 7 balls numbered from 1 to 7. If a ball is selected at random, and the number is noted down, and a second ball is selected without replacing the first, what is the probability that the two-digit number is even?
My Solution: if the number has to be even then its last digit is either 2,4,6.
Odd Set = {1,3,5,7}
Even Set = {2,4,6}
There are two possibilities to have our desired outcome:
I) Either both balls selected are even numbers therefore (C(1,3)*C(1,2)) / 7*6 = 6/42
II) The first one is odd but the other is even therefore (C(1,4)(1,3)) / 7*6 = 12/42
The result is : 6/42 + 12/42 = 18/42
Is this correct? Is there any easy way? Thank you