Question 856942
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Hi,
3 cards sum at most 9
1, 2, 3 
1, 2, 4
1, 2, 5
1, 2, 6
Select 3 <u>different</u> cards from Ten cards numbered from 1-10. 
P(sum &#8804; 9) = {{{ 4/(10C3 )}}} = 4/120 = 1/30