To solve this problem, we need to count the total number of ways that Umberto can remove two notes from his wallet, and then count the number of ways that he can remove two notes that add up to each of the given amounts.
There are a total of 4 possible notes that Umberto can choose from, so there are 4 choose 2 = 6 ways that he can remove two notes from his wallet.
To find the number of ways that he can remove two notes that add up to $11, we can observe that there is only one possible pair of notes: the $1 note and the $10 note. Therefore, the probability of getting a pair of notes that add up to $11 is 1/6.
To find the number of ways that he can remove two notes that add up to $70, we can observe that there is only one possible pair of notes: the $20 note and the $50 note. Therefore, the probability of getting a pair of notes that add up to $70 is 1/6.
To find the number of ways that he can remove two notes that add up to $80, we can observe that there is only one possible pair of notes: the $20 note and the $60 note (which is not in his wallet). Therefore, the probability of getting a pair of notes that add up to $80 is 0/6, which means it is impossible.
To find the number of ways that he can remove two notes that add up to at least $11, we can count the number of pairs of notes that add up to $11 or more. There are three such pairs: $1 and $10, $1 and $20, and $10 and $20. Therefore, the probability of getting a pair of notes that add up to at least $11 is 3/6, which simplifies to 1/2.
So, the probabilities for each part are:
A. $11: 1/6
B. $70: 1/6
C. $80: 0/6 (impossible)
D. at least $11: 1/2