Probability for: “Drawing $k$ red balls before the first blue one”

Question

Shortly I posted a similar question to this (here: Probability for "drawing balls from urn"). Actually, the following question was included in this topic as well::

In an urn there are $N$ balls, of which $N-2$ are red and the remaining are blue. Person $A$ draws $k$ balls, so that the first $k-1$ are red and the $k$ ball is blue. Then Person $B$ draws $m$ balls $\dots$:

What's the probability for Person $A$ to draw the first blue ball after drawing $k$ red ones? Meant is, that the drawing stops as soon as the blue ball has been drawn. So the first blue ball must be the last one for Person $A$

My ideas - following the answer from Jimmy R. (= direct approach):

1. $P(X=1) = \dfrac{2}{N}$
2. $P(X=2) = \dfrac{N-2}{N} \cdot \dfrac{2}{N-1}$
   $\dots$

Following this pattern I finally get

$$P(X=k) = \frac{\dbinom{N-2}{k}}{\dbinom{N}{k}} \cdot \frac{2}{N-k}$$

In fact, this solution differs only slightly from doing it via "Hypergeometric Distribution", which I did first. However, if I got it right then the Hypergeometric Distribution should be the wrong thing in this case. What remains is an explanation for Hypergeometric Distribution being not what I want, but here I stuck:

I'm citing wiki ("Hypergeometric distribution"): In probability theory and statistics, the hypergeometric distribution is a discrete probability distribution that describes the probability of k successes in n draws, without replacement, from a finite population of size N that contains exactly K successes, wherein each draw is either a success or a failure.

Let's see what I have here:

• $k$ successes in $n$ draws? Yes, if I see a success as "drawing blue ball" so I have $k=1$ success
• without replacement? Yes, since I won't put any ball back.
• a finite population of Size $N$? Yes, I have $N-2$ red ones and $2$ blue ones.
• exactly $K$ successes? Yes, in my case $K = 2$. I have two blue ones, so two successes.

So why won't this help me?

EDIT: Possible answer: It won't help, since I'm not interested in "$1$ success among $k$ draws" but in "$1$ success at the END of $k$ draws"?

Answer 1

The hypergeometric distribution can actually help, but in an indirect (and tricky and therefore preferably to be avoided) way if you think of this as follows: For any $k$ in $\{1,\dots, N-1\}$ you want to calculate the probability $P(X=k)$ where $X$ is the number of trials until the first success. Now fix $k$ and consider the following random variable

• $Y=$ number of successes (i.e. blue balls) in the first $k-1$ trials.

Now $Y$ indeed follows a hypergeometric distribution with parameters $N, K=2, n=k$ and therefore you can write $$P(X=k)=P(Y=0)\cdot\frac{2}{N-k}$$ In words: For a fixed $k$ you can express the probability of needing $k$ draws until you first success as the probability of $0$ successes in the first $k-1$ draws and success in the $k-$th draw. But be careful (the tricky part) that for each $k$ the random variable $Y$ will be different (since the parameters depend on $k$ by $n=k$ in the distribution of $Y$) so actually it is a $Y_k$.

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The pattern that you found from the direct approach is correct but note that it holds for $k\ge 2$. For $k=1$ the closed formula does not work, but that is a detail, you have it correct.

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Your possible answer in your edit seems valid. The point where the hypergeometric fails to solve this problem is in the first of your bullets. You actually do not have $1$ success in $n$ draws, but $1$ success at the end of $k$ draws. Moreover you are not interested in the number of successes but in the number of trials until the first success. Different things.