If 3% of electric bulbs manufacturing by a company are defective, the probability that a sample of 100 bulbs has no defective ones is 
 

Let the random variable X is the number of defective bulbs in a sample of 100 bulbs.

Given $\mathrm{\lambda }=3, \mathrm{r}=0$

$\mathrm{P}(\mathrm{x}=\mathrm{r})=\frac{{\mathrm{e}}^{-\mathrm{\lambda }}.{\mathrm{\lambda }}^{\mathrm{r}}}{\mathrm{r}!}$

Now the required probability 

$$\begin{gathered} \mathrm{P}(X=0)=\frac{{\mathrm{e}}^{-3}.3^0}{0!} \\ ={\mathrm{e}}^{-3} \end{gathered}$$