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