Step 1:

Let $X$ be the random variable denoting the number of defective fuses in a box of 200 fuses.

Probability of a defective fuse P=0.02

$\lambda=np$

$\quad=200\times 0.02$

$\quad=4$

$\therefore X\sim P(4) $

$\Rightarrow P(X=x)=\large\frac{e^{-4}4^4}{4!}$$\qquad x=0,1,2......$

Step 2:

Probability that more than 3 fuses are defective.

$P(X > 3)=1-P(X \leq 3)=1-e^{-4}[\large\frac{4^0}{0!}+\frac{4^1}{1!}+\frac{4^2}{2!}+\frac{4^3}{3!}]$

$\qquad\qquad=1-0.0183[\large\frac{(1+4+8)^3+32}{3}]$

$\qquad\qquad=1-0.4331$

$\qquad\qquad=0.5669$