The sum of two natural numbers n1 and n2 is known to be equal to 100.…

Total number of ways in which n₁ + n₂ = 100 is equal to 99.

Now, $n_1 . n_2>1600$

$\Rightarrow n_1\left(100-n_1\right)>1600$

$\Rightarrow n_1^2-100 n+1600<0$

$\Rightarrow \left(n_1-80\right)\left(n_1-20\right)<0$

$\Rightarrow 20<n_1<80$

$\Rightarrow 21 \leq n_1 \leq 79$

Thus, number of favorable ways

= 79 – 21 + 1 = 59

Hence, required probability = $\frac{59}{99}$