Here, we have four digit natural numbers, then total cases will exclude those number which contain zero at thousands place.
Hence, total cases will be
=(‌4C1×9×9×9)−(‌3C1×9×9)=2673 Again, only those numbers will have remainder 2 when divided by 5 either they have 2 at its unit place or 7 at its unit place.
When unit digit is 2 , then total number of four digit numbers will be
=(‌3C1×9×9)−(‌2C1×9)=225 When unit digit is 7 , then total number of four digit numbers will be
=8×9×9=648 Now, total favourable cases
=225+648=873 Required probability
=‌| ‌ Total favourable cases ‌ |
| ‌ Total number of cases ‌ |
=‌=‌