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Tina randomly selects two distinct
numbers from the set {1, 2, 3, 4, 5}, and
Sergio randomly selects a number from the set {1, 2, . . . , 10}. The probability that Sergio’s number is larger than the sum of the two numbers chosen by Tina is ?

the number of pairs that Tina could choose is C(5,2) = 10
For each of these, Sergio can choose 10 different numbers
So we have 100 possible ways for Tina to choose 2, and Sergio to choose one number.
We can actually list them
1 2 sum of digits = 3 leaving Sergio to choose 7 larger: prob = 7/100
1 3 sum of digits = 4 leaving Sergio to choose 6 larger: prob = 6/100
1 4 sum of digits = 5 leaving Sergio to choose 5 larger: prob = 5/100
1 5 sum of digits = 6 leaving Sergio to choose 4 larger: prob = 4/100
2 3 sum of digits = 5 leaving Sergio to choose 5 larger: prob = 5/100
2 4 sum of digits = 6 leaving Sergio to choose 4 larger: prob = 4/100
2 5 sum of digits = 7 leaving Sergio to choose 3 larger: prob = 3/100
3 4 sum of digits = 7 leaving Sergio to choose 3 larger: prob = 3/100
3 5 sum of digits = 8 leaving Sergio to choose 2 larger: prob = 2/100
4 5 sum of digits = 9 leaving Sergio to choose 1 larger: prob = 1/100

sum of those probs = 40/100 = 2/5

check my logic on this one.