**Toolbox:**

- \(In\;a\;batch\;of\;100\;watches\;10\;are\;defective\;\)
- \(p(a\;watch\;selected\;is\;defective)=\large\frac{10}{100}=\frac{1}{10}\)
- \(p=\large\frac{1}{10}\;q=1-\frac{1}{10}=\frac{9}{10}\)
- \(8\;watches\;are\;selected\;n=8\)
- \(p=(X=r)\;=\;\Large\;c^n_r\;p^r\;q^{n-r}\)

Answer:

\(p(at\;least\;1\;defective\;)=p(X\;\geq\;1)\)

=\(1-p(X\;<\;1)\)

=\(1-\;p(X\;=\;0)\)

\(=\Large\;1-c^8_o\;(\frac{1}{10})^0\;(\frac{9}{10})^8\)

\(=\Large\;1-(\frac{9}{10})^8\)