From hundred numbers 1 to 100, three distinct numbers are selected at random. Find theprobability that the selected numbers are divisible by 2 and 3.

To solve this problem, we first need to identify the numbers between 1 and 100 that are divisible by both 2 and 3. These numbers are also known as multiples of 6. They are: 6, 12, 18, ..., 96. As you can see, they increase by 6 each time. To find out how many there are, we can divide 96 (the largest multiple of 6 under 100) by 6. This gives us 16.

So, there are 16 numbers between 1 and 100 that are divisible by both 2 and 3.

Next, we need to calculate the total number of ways to choose 3 numbers from these 16. This is a combination problem, which can be solved using the combination formula: C(n, r) = n! / [r!(n-r)!], where n is the total number of items, r is the number of items to choose, and "!" denotes factorial.

So, the number of ways to choose 3 numbers from 16 is C(16, 3) = 16! / [3!(16-3)!] = 560.

Finally, we need to calculate the total number of ways to choose 3 numbers from 100, which is C(100, 3) = 100! / [3!(100-3)!] = 161700.

The probability that the 3 numbers chosen are divisible by both 2 and 3 is therefore 560 / 161700 = 0.00346, or approximately 0.35%.