<s>[INST] 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.....
 [/INST] 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 $2/10=20\%$
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