JEE Main 25 Feb 2021 Shift 2 Solved Paper

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 ‌

=‌

873

2673

=‌

97

297