If the integers X and (i are chosen at random between $1$ to $100$ then the probability that a number of the form $7^{\lambda }+ 7^{\mu }$ is dirisible by$5$ is

given $\lambda ,\mu €\left[1,100\right]$ and $7^{\lambda }+7^{\mu }$ should divisible by $5 .$

Now, the power of any positive numbers will have $4$ unit digits for different powers.

Exceptions are Number with units digit  0,6,15 raised to ower anything will have same units digits.

Numbers exceeds as 9 for only 2 different unit digits 9,1.

So for $7^{\lambda }+7^{\mu }$  to be divisible by $5$

Consider   $7^{4n+1}and n=0$  for $7^1$, we should have u of form 4m+3

 $\Rightarrow (3,7,....99)$   No of values of u $=\frac{99-3}{4}+1=25$

(We are dividing by h, as their difference  is $4$ and we are adding ‘1’ as we should include both $99$ and $3$)

$\therefore$ for every value of   there are $25$ values of u

$\therefore$ No of favorable cases$=25$

Now, total no of cases For every value of , there are 100 values of $‘u’$

$\therefore$ Probability $=1/4$