what is wrong in my answer and what will be the correct solution for this probability question ? A and B play a game where each is asked to select a number from 1 to 5. If 
the two numbers match, both of them win a prize. The probability that they 
will not win a prize in a single trial is - 
a)1/25 b) 24/25 c)2/25 d)23/25 
My proposed solution - Total number of ways in which both of them can select a number each:
=5×5=25
total number of ways in which two numbers match that is (1,1),(2,2),(3,3),(4,4),(5,5) = 5
probability that they will not win a prize = 1 - 5/25  = 4/5.
Tell me what is wrong in my answer and what will be the correct solution ? 

The rationale behind the solution is this:
The game conductor has a particular number from 1 to 5 that he wants A and B to pick.  Let us say "2". Then
P(A picking 2) $= \frac{1}{5}$
P(B picking 2) $= \frac{1}{5}$
P(A picking other than 2) $= \frac{4}{5}$
P(B picking other than 2) $= \frac{4}{5}$
P(Both picking 2) $= \frac{1}{5}\cdot\frac{1}{5}$
P(A picks 2, B picks other than 2)$ = \frac{1}{5}\cdot\frac{4}{5}$
P(A picks other than 2, B picks 2) $= \frac{4}{5}\cdot\frac{1}{5}$
P(B picks other than 2, B picks other than 2)$ = \frac{4}{5}\cdot\frac{4}{5}$
They don't win the game when the last three scenarios happen $= \frac{1}{5}\cdot\frac{4}{5}+\frac{4}{5}\cdot\frac{1}{5}+\frac{4}{5}\cdot\frac{4}{5} = \frac{24}{25}$
They win the prize $= \frac{1}{5}\cdot\frac{1}{5} = \frac{1}{25}$
Obviously, the question has been worded badly.  Your answer is correct for the wording of the question.  
Thanks
Satish
