\(P=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{99}\right)\\ =\left(\dfrac{1}{2}\right)\left(\dfrac{2}{3}\right)\left(\dfrac{3}{4}\right)...\left(\dfrac{98}{99}\right)\\ =\dfrac{1\cdot2\cdot3\cdot....\cdot98}{2\cdot3\cdot4\cdot...\cdot99}\\ =\dfrac{99}{101}\)