A probability question: Poor Alex 
Alex remembers all but the last digit of his friend's telephone number. He decides to choose the last digit at random in an attempt to reach him. Given that, Alex has only enough money to make two phone calls, the probability that he dials the right number before running out of money can be expressed as an irreducible fraction p/q.

What I did was--

WLOG let us assume the correct number be 1. Total number of outcomes can be calculated this way
(0,1)(0,2)(0,3)(0,4)(0,5)(0,6)(0,7)(0,8)(0,9)

(1)

(2,0)(2,1)(2,3)(2,4)(2,5)(2,6)(2,7)(2,8)(2,9)
(3,0)(3,1)(3,2)(3,4)(3,5)(3,6)(3,7)(3,8)(3,9)
(4,0)(4,1)(4,2)(4,3)(4,5)(4,6)(4,7)(4,8)(4,9)
(5,0)(5,1)(5,2)(5,3)(5,4)(5,6)(5,7)(5,8)(5,9)
(6,0)(5,1)(5,3)(5,4)(5,6)(5,7)(5,7)(5,8)(5,9)
(7,0)(7,1)(7,2)(7,3)(7,4)(7,5)(7,6)(7,8)(7,9)
(8,0)(8,1)(8,2)(8,3)(8,4)(8,5)(8,6)(8,7)(8,9)
(9,0)(9,1)(9,2)(9,3)(9,4)(9,5)(9,6)(9,7)(9,8)
So total number of outcomes is 82 and total number of favorable outcomes is 10 so my answer is 10/82 or 5/41. But it is not correct. Please help me understand this.....

Alex tries two numbers. Then the probability that the correct number belongs to this pair is simply 
$$\frac{2}{10} = 20 \%$$
