What is the probability of selecting all the blue balls?

Question

We have a bag full of many balls (let's say n balls). We cannot see inside of the bag but we are certain that only a certain number of them are blue (let say r). If we want to randomly choose half of the total balls in the bag, what is the probability that we have selected all of the blue balls in our sample?

Here is how I think of this. If the number of balls in the bag are much larger than the number of blue balls. I can think of all blue balls as one single package and group other balls in groups with the same size. Then the prroblem reduces to: what is the probablity of selecting the blue package among the selection of half of the total groups which is equal to:

$${(m-1)\over ^m C_{m/2}} = {{m\over 2}! {m\over 2}!\over m \times (m-2)!}$$

where m is the number of groups. This converges to the exact solutiuon when m is big. But I'm still uncertain about the exact solutiuon of this problem. I appreciate you sharing your thoughts and comments.

Answer 1

This can be reduced to a combinatorics problem. Let's make it a little more general.

Suppose you have $n$ balls, of which $r$ are blue. You select $k,$ where $k > r.$

What are the total number of ways you can select $k$ balls out of $n?$

Then, what are the total number of ways you can select $k$ balls out of $n,$ where $r$ of the $k$ balls are necessarily the blue balls?

Divide the latter answer by the former, and you have your answer.

CLARIFICATION: The answer is easy if in the second part you think of all blue balls as a single package. Then to the latter point, suppose we have already selected this blue package. What is the other possible combinations? The answer would be choosing k-r (not k, because we have already selected all r blue balls and put it aside) from n-r remaining balls. In summary the probability that we have selected all of the blue balls in our k sample is equal to

$${^{n-r}C_{k-r}\over ^n C_k} = {(n-r)! k! \over (k-r)! n!}$$. For the special case of k=n/2 we have

$${^{n-r}C_{{n\over 2}-r}\over ^n C_{n\over 2}} = {(n-r)!{n\over 2}!\over ({n\over 2} -r)! n!}$$

Answer 2

It's a typical brain teaser. Invert the question: what is the probability that none of the blue balls are scooped out of the bag?

All of a sudden it's easy to answer! $$\frac{n-r}{n}\frac{n-r-1}{n-1}\dots\frac{n-m+1-r}{n-m+1}=\frac{(n-m)!(n-r)!}{n!(n-r-m)!}$$

You pull the first ball, and it's not blue. How likely it is? $\frac{n-r}{n}$ So, you keep pulling them until you got $m$ balls out, and none of the blue balls showed up yet. And you get the answer above, which is also the answer to the original question where $m=n/2$. thanks to @Bridgeburners comment

Here's why this works. The original problem is formulated in terms of the balls that were pulled (chosen) from the bag: we want all the blue balls out. It's shown in the picture below.

However, the next picture shows that the answer got to be the same in terms of the inverted problem about all balls being red, i.e. none of the balls being blue. So, if the question is about probability of all $r$ blue balls being chosen in the group of $k$ balls, then it's equivalent to a question about probability of $m=n-k$ balls being left in the bag after $k$ balls are removed from it. Hence, in terms of the $k$ balls out the same answer can be written as: $$\frac{k!(n-r)!}{n!(k-r)!}$$

To summarize, the solution is very simple and intuitive if you re-formulate the problem in terms of the unchosen balls being all red.