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
Best Answer
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 $\frac{18}{90} = \frac{1}{5}$.