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.
Best Answer
Here lies your problem:
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
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 $$\mathcal{P}(A) = \mathcal{P}(A_1)+\mathcal{P}(A_2)+\mathcal{P}(A_3) = \frac{\binom{4}{1}\cdot\binom{16}{2}+\binom{4}{2}\cdot\binom{16}{1}+\binom{4}{3}\cdot\binom{16}{0}}{\binom{20}{3}}$$