<s>[INST] 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. 
 [/INST] 
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 
$$\binom{1}{1}\binom{4}{1}\binom{16}{2} + \binom{2}{1}\binom{4}{2}\binom{16}{1} + \binom{3}{1}\binom{4}{3}\binom{16}{0} = \binom{4}{1}\binom{19}{2}$$
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