Question: Suppose that you buy a lottery ticket containing k distinct numbers from among ${1, 2, . . . , n}$ where $1 ≤ k ≤ n$. To determine the winning tickets, k balls are randomly drawn without replacement from a bin containing n balls numbered $1, 2, . . . , n$.
(a) What is the probability that at least one of the numbers on your lottery ticket is among those drawn from the bin?
(b) Prizes start when you have at least $\frac{k}{2}$ (ie. half) numbers in common with the drawn balls. Express the probability that you win something. You may find it helpful to use summation notation.
*For this question, I got question (a) done as I got $(1 - \frac{((n-k)!)2}{n!(n-2k)!})$. But I am confused on how to do question (b). Can anyone help me out please?
The chance of exactly $r$ hits : $r \in \{0,1,2, \cdots, k\}$ is
$$F(r) = \frac{\binom{k}{r}\binom{n-k}{k-r}}{\binom{n}{k}}.$$
For part (a)
$$1 - F(0) = 1 - \frac{\binom{k}{0}\binom{n-k}{k}}{\binom{n}{k}} = 1 - \frac{[(n-k)!]^2}{(n!)[(n-2k)!]}.$$
For part (b)
Let $s$ denote the smallest positive integer greater than or equal to $\frac{k}{2}.$
Chance of winning something is
$$\sum_{r=s}^k F(r).$$
Edit
It has been suggested that I edit my answer to part (b) by eliminating use of the variable $s$. I would like to explain why I have rejected this suggestion.
As stated, $k$ may be an odd #. When $k$ is odd, you have to have at least $\frac{k+1}{2} = s$ numbers in common with the drawn balls to win a prize.
Suppose (for example) that $k = 7$. I am unaware of any universally accepted interpretation of (for example) $\left[\sum_{r = (7/2)}^7 F(r)\right]$ as signifying
$F(4) + F(5) + F(6) + F(7).$
Even if there is such a generally accepted interpretation, I was not aware of it, and I think that many readers of this response are not aware of it. Therefore, even if there is such a generally accepted interpretation, I prefer using the $s$ variable so that my answer will be crystal clear.