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.....
 A: So you are choosing twice from a set of ten Without replacement. That means the first digit tried, called it A is taken from 0 through 9. The second digit, digit B, should be different, because A has been eliminated. 
Let's say the correct digit is X. 
What are the chances that neither A or B is X? 
That question can be restated: How many ways are there to pick two digits from nine (the ten digits minus X), divided by the number of ways of picking two digits. 
There are 36 ways of picking a combination of two distinct digits none of which is X. 
There are 45 ways of choosing two distinct digits from all 10. 
The probability of failing to get the right digit is 4/5. 
So the probability of success is 1/5=1-4/5. 
Alternatively: 
There's a 1/10 chance of getting it right on the first try. 
There's a 9/10 chance of not. We only make the second try if we failed the first. The chances of succeeding the second time around is is 1/9. 
So the chances of success is (1/10)(1/1)+(9/10)(1/9)=2/10=1/5, just as via the other method. 
This is all based on choosing two distinct digits. If you repeat the same failed digit, the analysis changes a little. 
A: Alex tries the first number. The probability to get it wrong is $9/10$. If he does, he chooses the second number. The probability that he get the second number wrong is $8/9$ (assuming that he does not try to dial the same number expecting a different result). Then the total probability that he gets the number wrong is $$\frac 9{10}\frac89=\frac8{10}$$
That means that he get's it right with probability $2/10=20\%$
A: Alex tries two numbers. Then the probability that the correct number belongs to this pair is simply 
$$\frac{2}{10} = 20 \%$$
