Probability that if three balls numbered from 1-20 are selected without replacement that at least one will be numbered at least 17 I'm a new user here. I am currently self-learning probability theory (long journey ahead). I was wondering as to why my answer is not correct. I have listed the question below.
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 as large as or larger than 17, what is the probability that we win the bet?
My solution: $ {{ 4 \choose 1}{ 19 \choose 2} \over { 20 \choose 3}}  $
Reason: ${ 4 \choose 1}$ is of choosing either 17, 18, 19, or 20. ${ 19 \choose 2}$ once a urn is chosen which is either 17,18, 19 ,20. There are 19 balls left and the value does not matter. 
 A: Here lies your problem:

${19 \choose 2}$ once a urn is chosen which is either 17,18, 19 ,20.
  There are 19 balls left and the value does not matter.

You are not partioning the desired set correctly. The selection for the remaining ball should be made out of the sixteen balls labelled $\{1,2,\dots,16\}$. In addition the problem says 

"we bet that at least one of the balls that are drawn has a number as
  large as or larger than 17"

so after making the above correction you should count the events where $2$ of the balls and $3$ of the balls belong to $\{17,18,19,20\}$ like so
Solution. Let $\mathcal{P}(A)$ denote the required probability and Define $A_i$ to be the event that $i$ of $3$ the balls have a value of at least $17$ for $i\in\{1,2,3\}$ then
$$\mathcal{P}(A) = \mathcal{P}(A_1)+\mathcal{P}(A_2)+\mathcal{P}(A_3) = \frac{\binom{4}{1}\cdot\binom{16}{2}+\binom{4}{2}\cdot\binom{16}{1}+\binom{4}{3}\cdot\binom{16}{0}}{\binom{20}{3}}$$
A: 
Find the probability that at least one of three balls selected from an urn containing $20$ balls, numbered from $1$ to $20$, displays a number that is at least $17$.

Method 1:  We subtract the probability that none of the three selected balls have numbers larger than $16$ from $1$.  
There are 
$$\binom{20}{3}$$
ways to select three of the $20$ balls.
There are 
$$\binom{16}{3}$$
ways to select three of the $16$ balls with numbers less than $17$.  
Hence, the probability that none of the three balls have a number that is at least $17$ is 
$$\frac{\dbinom{16}{3}}{\dbinom{20}{3}}$$
Therefore, the probability that at least one ball has a number that is at least $17$ is 
$$1 - \frac{\dbinom{16}{3}}{\dbinom{20}{3}}$$
Method 2:  We count directly.
There are four balls with numbers at least $17$ in the urn and $20 - 4 = 16$ balls with smaller numbers.  The number of ways of selecting exactly $k$ balls with numbers that are at least $17$ and $3 - k$ balls with numbers less than $17$ is 
$$\binom{4}{k}\binom{16}{3 - k}$$
If at least one ball with a number at least $17$ is selected, then $k \geq 1$, so the number of favorable cases is 
$$\binom{4}{1}\binom{16}{2} + \binom{4}{2}\binom{16}{1} + \binom{4}{3}\binom{16}{0}$$
The probability that at least one ball that displays a number at least $17$ is thus 
$$\frac{\dbinom{4}{1}\dbinom{16}{2} + \dbinom{4}{2}\dbinom{16}{1} + \dbinom{4}{3}\dbinom{16}{0}}{\dbinom{20}{3}}$$
What was your mistake?
In designating one of the balls to be the one that displays a number that is at least $17$, you counted selections in which two balls display a number that is at least $17$ twice and selections in which all three balls display a number that is at least $17$ three times, once for each way you could designate one of those balls as the one that displays a number that is at least $17$.  Notice that 
$$\binom{1}{1}\binom{4}{1}\binom{16}{2} + \binom{2}{1}\binom{4}{2}\binom{16}{1} + \binom{3}{1}\binom{4}{3}\binom{16}{0} = \binom{4}{1}\binom{19}{2}$$
