Xin phép sửa lại đề.
\(P=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right)\left(1-\frac{1}{100}\right)\)
\(1-\frac{1}{2}=\frac{1}{2}\)
\(1-\frac{1}{3}=\frac{2}{3}\)
\(1-\frac{1}{4}=\frac{3}{4}\)
.........................
\(1-\frac{1}{99}=\frac{98}{99}\)
\(1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow P=\frac{1.2.3...98.99}{2.3.4...99.10}\)
\(P=\frac{\left(1.2.3...98.99\right)}{\left(2.3.4...99.100\right)}\)
\(P=\frac{1}{100}\)
Vậy: P = 1/100