\(P=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right)\)
\(P=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{98}{99}\)
\(P=\frac{1.2.3...98}{2.3.4...99}=\frac{1}{99}\)
Ta có: \(P=\left\{1-\frac{1}{2}\right\}.\left\{1-\frac{1}{3}\right\}....\left\{1-\frac{1}{99}\right\}\)
\(\Rightarrow P=\frac{1}{2}.\frac{2}{3}.........\frac{98}{99}\)\(=\frac{1.2.3...98}{2.3.4...99}=\frac{1}{99}\)
Vậy \(P=\frac{1}{99}\)