Probability of getting 7 numbers smaller or equal to 59, from a draw of 20 numbers out of 80

Question

What is the Probability of getting $7$ numbers smaller or equal to $59$, from a draw of $20$ numbers out of $80$?

I can work out the probability of getting them in the 1st $7$ numbers

$$59/80 * 58/79 * ... * 53/74 = 0,107390589$$

However, I see there are many other possibilities. Getting $7$ numbers under $59$ from the first $8$ numbers, from the first $9$ numbers, ... etc.

Should I calculate them one by one from $7$ to $20$ numbers, and them add them all?

Thanks!

Answer 1

The probability of getting $n$ numbers smaller or equal to $59$ by choosing $20$ numbers from $\{1,2,3,...,80\}$ is like the probability of choosing $n$ numbers from $\{1,2,3,...,59\}$, and the rest of them ($20-n$ numbers) from $\{60,61,...,80\}$ with all legible values of $n$, thus:

$\displaystyle p_n=\dfrac{{59 \choose n}{80-59 \choose 20-n}}{{80 \choose 20}}=\dfrac{{59 \choose n}{21 \choose 20-n}}{{80 \choose 20}}$

As we want at least $7$ numbers less than or equal to $59$ we should say $7 \leq n \leq 20$. Hence, the probability will be:

$\displaystyle \sum_{n=7}^{20} p_n=\sum_{n=7}^{20} \dfrac{{59 \choose n}{21 \choose 20-n}}{{80 \choose 20}}$

Answer 2

Kind of a stupid method but i think its easy to understand why it works just by looking at it:

let $a = \frac{59*58*...*53}{80*79*...*74}$ then $$a + \sum^{x = 12}_{x=1}{\Big(aC_{x}^{7+x}\prod^{n=x}_{n=1}{\frac{22-n}{74-n}}\Big)}$$