A box contains $3$ red balls, $4$ blue balls, $6$ green balls. Balls are drawn one-by-one without replacement until all the red balls are drawn. Let $D$ be the number of draws made. Calculate $P(D \le 9)$.
Since we are drawing without replacement, this is a hypergeometric distribution.
Minimum of $D$ is $3$.
$P(D \le 9) = \sum_{i=3}^{9}\frac{\binom{3}{3}\binom{10}{i-3}}{\binom{13}{i}}=0.734265734$
But the answer is just
$\frac{\binom{3}{3}\binom{10}{9-3}}{\binom{13}{9}}=0.293706293$
Why does it only consider the case $P(D=9)$?
Edit 1: Pictures of the question and solution from the textbook
Best Answer
How should we interpret $$\frac{\binom{3}{3}\binom{10}{9-3}}{\binom{13}{9}}$$ as a probability of drawing balls from the box?
The denominator, $\binom{13}{9},$ counts all possible ways to select a set of $9$ items from $13$ distinguishable items. So merely by using this term, we have already assumed we have some way of distinguishing the balls other than color (even if it's only properties such as "the ball that was in the rear right corner of the box" or "the ball that was originally $3$ cm from the rear wall and $9$ cm from the left wall").
To put it another way, if we consider every set of $9$ balls containing $2$ red balls, $3$ blue, and $4$ green to be the same outcome, there are only $(3+1)(4+1)(6+1) = 140$ possible outcomes. But $\binom{13}{9}=715,$ so it can't possibly be counting a subset of those $140$ outcomes.
In order to make sense of this way of counting, let's arbitrarily stick numbers on the balls so we can tell which is which: $$r_1,r_2,r_3,b_1,b_2,b_3,b_4,g_1,g_2,g_3,g_4,g_5,g_6.$$ Now we can identify $\binom{13}{9}$ different ways to draw nine balls.
The denominator $\binom{13}{9}$ does not distinguish sequences in which the balls could be drawn out, so it should treat the sequence of draws $r_1,r_2,b_1,b_2,b_3,b_4,g_1,r_3,g_2$ the same as the sequence $r_1,r_2,b_1,b_2,b_3,b_4,g_1,g_2,r_3.$ (After all, $\binom{13}{9}$ is also the number of outcomes in a problem in which the first step is to draw $9$ balls without stopping.) But in the first case, we would never actually draw $g_2$ (according to the problem statement) since we had already drawn all three red balls.
Now how do we make sense of that? One way is to suppose that the balls are already lined up inside the box in some random sequence in which they are waiting to be drawn, like a deck of cards. And now we can say that $\binom{13}{9}$ is exactly the number of ways to choose which subset of the original $13$ balls is found in the first $9$ places in the drawing sequence, even in cases where the drawing stops before all the balls have been drawn.
One of the $\binom{13}{9}$ subsets of balls is the subset $\{r_1,r_2,r_3,b_1,b_2,b_3,b_4,g_1,g_2\}.$ These balls might be lined up in the sequence $r_1,r_2,b_1,b_2,b_3,b_4,g_1,g_2,r_3$ (resulting in $D=9$), but they could just as well occur in the sequence $r_1,r_2,r_3,b_1,b_2,b_3,b_4,g_1,g_2$ (resulting in $D=3$). By distinguishing only $\binom{13}{9}$ subsets of balls to make the denominator of our probability, we have limited what we know about each of the $\binom{13}{9}$ events counted by that denominator; in fact, the only useful thing we can say about each event (as far as this problem is concerned) is whether the subset contains all of the red balls (in which case we know only that $D\leq 9$) or that the subset does not contain all the red balls (in which case $D > 9$).
The numerator $\binom{3}{3}\binom{10}{9-3}$ counts the subsets of $9$ balls that contain all of the red balls. Therefore it tells us when $D \leq 9,$ and $$\frac{\binom{3}{3}\binom{10}{9-3}}{\binom{13}{9}} = P(D \leq 9).$$
To get the probability of exactly nine draws, $P(D=9),$ I still think the easiest method is to take all the cases in which there are nine or fewer draws, and subtract all the cases in which there are fewer than nine draws. That is, $$ P(D=9) = P(D\leq 9) - P(D\leq 8). $$ But if you really want to compute $P(D=9)$ directly, you can do it this way:
First you must draw exactly two red balls in the first eight draws, and then you must draw the third red ball on the next draw. The probability of exactly $2$ red balls and $6$ others in the first $8$ draws is $$ \frac{\binom{3}{2}\binom{10}{6}}{\binom{13}{8}}. $$ There are then just five balls remaining, of which exactly one is red, and the probability that the very next draw will be that red ball is therefore $\frac15.$ We therefore have $$ P(D=9) = \frac15\cdot \frac{\binom{3}{2}\binom{10}{6}}{\binom{13}{8}}. $$
You might find it instructive to verify that this gives the same result as the subtractive method, that is, that $$ \frac15\cdot \frac{\binom{3}{2}\binom{10}{6}}{\binom{13}{8}} = \frac{\binom{3}{3}\binom{10}{9-3}}{\binom{13}{9}} - \frac{\binom{3}{3}\binom{10}{8-3}}{\binom{13}{8}}. $$