$\begingroup$I am trying to figure out how to set up this problem or any similar one. If you have 10 balls numbered from 1 to 10, and you pick 5 balls, what is the probability that you will have picked ball#1? In other words, how many possible combinations of 5 numbers are there in the set of numbers (1-10), without repeating any number, and how many of those combinations include the number 1?
$\endgroup$13 Answers
$\begingroup$Imagine that you pick the balls and put them in 5 slots. If you want to have the ball #1 in one of these slots, you simply occupy one of them with this, and you need to draw the other four balls. You have $ 9 \choose 4$ possibilities for the other four balls. at this point you can mix the balls in the boxes, hence I would say you have: $ 5!$ $9 \choose 4 $ combinations.
$\endgroup$2$\begingroup$You're picking half the balls out of all combinations. The probability you pick ball #1 is 50%. In other words:
All possible combinations of 5 balls out of 10: $\binom{10}{5}=252 $. All combinations with the ball #1: $\binom{9}{4}=126$. This is because we take every other ball and make all the possible combinations of 4 balls with them. You just stick the ball #1 at the end of each one of those. So, $\frac{126}{252}=0.5$.
As noted by JMoravitz, if order matters then you calculate the permutations of each possible combination, and since it's the same for every combination, $5!$, you multiply by all combinations. That is, $5!\binom{10}{5}$ in total, and $5!\binom{9}{4}$ with the ball #1. Still, $\frac{5!\binom{9}{4}}{5!\binom{10}{5}}=\frac12$.
$\endgroup$1$\begingroup$The number of ways of selecting a subset of $k$ elements from a set with $n$ elements (selecting $k$ of the $n$ elements without regard to order and without repetition of elements) is $$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$ Therefore, the number of ways to select five of the ten balls is equal to the number of five-element subsets of a set with ten elements, which is $$\binom{10}{5}$$ A selection of five balls that includes ball number $1$ requires the selection of ball number $1$ and four of the other nine balls. Such selections can be made in $$\binom{1}{1}\binom{9}{4} = \binom{9}{4}$$ ways. Hence, the probability that a selection of five of the ten balls will include ball number $1$ is $$\frac{\dbinom{9}{4}}{\dbinom{10}{5}}$$
$\endgroup$
