If the integers m and n are chosen at random between 1 and 100, then the probability that a number of the form 7^m+ 7ⁿ is divisible by 5, is
equal to

$7^1=7,7^2=49,7^3=343,7^4=2401,\dots$

Therefore, for $7^{\mathrm{r}},\mathrm{r}\in \mathrm{N}$ the number ends at unit place 7,9, 3, 1, 7,
$\therefore 7^{\mathrm{m}}+7^{\mathrm{n}}$ will be divisible by 5, if it ends with 5 or 0.
But it cannot end at 5.
Also, it cannot end at 0.
For this m and n should be as follows

For any given value of m, there will be 25 values of n,
Hence, the probability of the required event is $\frac{100\times 25}{100\times 100}=\frac{1}{4}$