Question 368260
The experiment follows a hypergeometric distribution:  there are 5 good fuses and 5 bad fuses.
1st case, 1 bad and 2 good fuses:  # of ways is C(5,1)*C(5,2) = 50.
2nd case, 2 bad and 1 good fuse:  # of ways is C(5,2)*C(5,1) = 50.
3rd case, 3 bad and 0 good fuse: # of ways is C(5,3)*C(5,0) = 10.
The number of ways of choosing any 3 out of 10 fuses in a box is C(10,3) = 120.
Therefore the probability is (50+50+10)/120 = 110/120 = 11/12.