Probability of choosing positive 2-digit integer with 4 in either place GRE study exam guide has following


  If an integer is randomly selected from all positive 2-digit integers,
  what is the probability that the integer chosen has at least one 4 in
  the tens place or the units place?


I understand probability of being in 10s place is $1/9$ and the probability of being in the units place is $1/10$

When I add $1/9$ + $1/10$ the answer is $19/90$. However, answer says $1/5$.

Please explain
A = $4$ in a ten place,
B = $4$ in a unit place,
C = at least one $4$ in a ten or unit place.

$P(A) = \frac{1}{9}$, $P(B) = \frac{1}{10}$, $P(C) = \frac{1}{9} + (1-\frac{1}{9})*\frac{1}{10} = \frac{10 + 8}{90} = \frac{1}{5}$.

Here you go:
You have $10$ cases for event $A$: $40, 41, 42, 43,44,45,46,47,48,49$, and $9$ cases for event $B$: $14, 24, 34, 44, 45, 46, 47, 48, 49$. So when you get $A \cup B$ you have not $19$ cases, but 18. There are $90$ two-digit numbers, so $\frac{18}{90} = \frac{1}{5}$.
P(AorB)=P(A) +P(B) _P(both A and B)

P(both A and B)= P(A) * P(B) events are not mutually exclusive..
So P(A)= 1/9, P(B) = 9/90
               = 1/9 + 9/90 _ 1/9*9/90= 1/5 answer..