Yes I botched this one :-) (My face is red!)
However the other tutor botched it too, in another way.
The odds in favor are
number of successful choices : number of unsuccessful choices
or
81:62
(not 81:100).
Edwin
Answer by sgudihal(1) (Show Source): You can put this solution on YOUR website! In the solution given above,
we have to include other numbers like 51, 52,53, 54, 55(already included by tutor), 56, 57, 58 and 59 apart from the 10 mentioned
So that makes it 19 and the odds then are
81:100