What is the probability of getting a number between 1 and 100 which is divisible by 1 and itself only ? (itmeans prime nos. but then how do you find all theprime nos. all the way from 1 to 100.....do u sit and go through all the nos one by one? )

The prime numbers between 1 and 100 are :
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.
So the number of prime numbers are 25
Number of favourable outcomes = 25
There are total of 98 numbers between 1 and 100.
So, total number of possible outcomes = 98

$\mathrm{p}\left(\mathrm{gettin} \mathrm{a} \mathrm{prime} \mathrm{number}\right) =\frac{\mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{total} \mathrm{possible} \mathrm{outcomes}} =\frac{25}{98}$