Three balls are to be randomly selected without replacement from an urn containi

Question

Three balls are to be randomly selected without replacement from an urn containing 20 balls numbered 1 through 20. If we bet that at least one of the balls that are drawn has a number no less than 17, what is the probability that we win the bet? I already have the solution and the setup for the solution looks like attached image:

My question is that in the numerator where did x-1 and 2 come from? I am clear on the denominator but lets say we want to figure out the probability of at least 1 ball to be =>17 so that leaves us two balls.. but where is x-1 coming from? is x total number of balls and I am reducing it by 1?

Explanation / Answer

There are 2 ways to solve this problem. One uses conditional expectation and the other counts the number of ways something can be done.

With both methods, we will be using the fact that the probability of selecting at least one ball greater than or equal to 17 = 1 - P(selecting 3 balls 16 or less), and will actually calculate P(selecting 3 balls 16 or less)

Method 1, using conditional expectation.

P(first ball is 16 or less) = 16/20 (16 balls less than or equal to 16/20 balls less than or equal to 20)

P(second ball is 16 or less|first ball is 16 or less) = 15/19 (since the first ball is less than or equal to 16, there are only 15 balls remaining that are less than or equal to 16; as one ball has been selected, there are 19 balls left).
P(third ball is 16 or less|first and second balls or less than or equal to 16) = 14/18 (as 2 balls less than or equal to 16 have been selected, only 14 remain that are less than or equal to 16; meanwhile, as 2 of the 20 balls have been selected, 18 balls remain).
Then, P(first, second, and third balls are less than or equal to 16) =
P(first ball is 16 or less)P(second ball is 16 or less|first ball is 16 or less)P(third ball is 16 or less|first and second balls or less than or equal to 16) =
16/20*15/19*14/18 = 28/57
Then, P(at least one ball is greater than or equal to 17) = 1 - 28/57 = 29/57 0.508771929824561

Method 2.

The number of ways that 3 balls can be selected out of 16 is C(16,3)= 16*15*14/3!

The number of ways that 3 balls can be selected out of 20 is C(20,3) = 20*19*18/3!

Then, the probability that the 3 balls selected from 1 -20 are between 1 and 16 is

C(16,3)/C(20,3) =  16*15*14/3!/(20*19*18/3!) =

16*15*14/(20*19*18) = 28/57

Then, P(at least one ball is greater than or equal to 17) = 1 - 28/57 = 29/57 0.508771929824561

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