Probability Part 1

First Case: Choose two non-zero digits ‌9C2
Now, number of 5 -digit numbers containing both digits =25−2
Second Case: Choose one non-zero & one zero as digit ‌9C1.
Number of 5 -digit numbers containg one non zero and one zero both =(24−1)
Required prob.
=

(‌9C2×(25−2)+‌9C1×(24−1))

9×104

=

36×(32−2)+9×(16−1)

9×104

=

4×30+15

104

=

135

104