· Case III:I3>L1+L2 P(no triangle via case III)= by symmetry with case I Since the three cases are mutually exclusive, the probability that L1, L2, and L cannot form a triangle is 4 +i+5=3. Therefore the probability that the three pieces of the stick form a triangle is1-是=1• Case III: L3 > L1 + L2 P(no triangle via case III)= 1 4 by symmetry with case I. Since the three cases are mutually exclusive, the probability that L1, L2, and L3 cannot form a triangle is 1 4 + 1 4 + 1 4 = 3 4 . Therefore the probability that the three pieces of the stick form a triangle is 1 − 3 4 = 1 4 . 2