Posted by maria on Thursday, November 24, 2011 at 8:30am.
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.
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