Given: Total number of watches = \[100\] and number of defective watches = \[10\]
So, the probability of selecting a detective watch = \[10/100\text{ }=\text{ }1/10\]
Now,
n = \[8\], \[p\text{ }=\text{ }1/10\]and \[q\text{ }=\text{ }1\text{ }\text{ }1/10\text{ }=\text{ }9/10,\text{ }r\text{ }\ge \text{ }1\]
\[P\left( X\text{ }\ge \text{ }1 \right)\text{ }=\text{ }1\text{ }\text{ }P\left( x\text{ }=\text{ }0 \right)\text{ }=\text{ }1\text{ }{{~}^{8}}{{C}_{0}}~{{\left( 1/10 \right)}^{0}}{{\left( 9/10 \right)}^{8\text{ }\text{ }0}}~=\text{ }1\text{ }\text{ }{{\left( 9/10 \right)}^{8}}\]
Therefore, the required probability is \[1\text{ }\text{ }{{\left( 9/10 \right)}^{8}}\].