From the set of 10 numbers {1, 2,...,10} three numbers are selected at random without replacement. The probability that the sum of these selected numbers is 9, is (A) 1/40 (B) 1/20 (C) 3/10 (D) 3/80

To solve this problem, we first need to find all the possible combinations of 3 numbers from the set {1, 2,...,10} that add up to 9.

The combinations are: {1, 2, 6}, {1, 3, 5}, {2, 3, 4}. So, there are 3 favorable outcomes.

Next, we need to find the total number of ways to select 3 numbers from the set {1, 2,...,10} without replacement. This is a combination problem, and the formula for combinations is nCr = 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 total number of ways to select 3 numbers from 10 is 10C3 = 10! / 3!(10-3)! = 120.

Therefore, the probability that the sum of the selected numbers is 9 is the number of favorable outcomes divided by the total number of outcomes, which is 3/120 = 1/40.

So, the answer is (A) 1/40.