Let’s assume X to be the random variable denoting a bulb to be defective.
Here, \[n\text{ }=\text{ }10,\text{ }p\text{ }=\text{ }1/50,\text{ }q\text{ }=\text{ }1\text{ }\text{ }1/50\text{ }=\text{ }49/50\]
We know that, \[P\left( X\text{ }=\text{ }r \right)\text{ }={{~}^{n}}{{C}_{r}}~{{p}^{r~}}{{q}^{n\text{ }\text{ }r}}\]
(i) None of the bulbs is defective, i.e., \[r=0\]
\[P\left( x\text{ }=\text{ }0 \right)\text{ }={{~}^{10}}{{C}_{0}}~{{\left( 1/50 \right)}^{0}}{{\left( 49/50 \right)}^{10\text{ }\text{ }0}}~=\text{ }{{\left( 49/50 \right)}^{10}}\]
(ii) Exactly two bulbs are defective
So, \[P\left( x\text{ }=\text{ }2 \right)\text{ }={{~}^{10}}{{C}_{2}}~{{\left( 1/50 \right)}^{2}}{{\left( 49/50 \right)}^{10\text{ }\text{ }2}}\]
= \[{{45.49}^{8}}/{{50}^{10}}~=\text{ }45\text{ }\times \text{ }{{\left( 1/50 \right)}^{10}}~\times \text{ }{{49}^{8}}\]
(iii) More than \[8\] bulbs work properly
We can say that less than \[2\] bulbs are defective
\[P\left( x\text{ }<\text{ }2 \right)\text{ }=\text{ }P\left( x\text{ }=\text{ }0 \right)\text{ }+\text{ }P\left( x\text{ }=\text{ }1 \right)\]
