Drawing Balls Without Replacement 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
It's just a counting argument.

Choose the three red balls, and choose six others. Divide by the total number of possible ways to draw nine balls.

When you get the three red balls doesn't matter, as long as they're part of the nine. You just keep drawing until you get nine balls if you already have the three red ones.

Or, just grab nine balls from the box all at once, if that's conceptually easier.
You're doing a drawing balls without replacement, what conduct us to supose that is hypergeometric distribution as follows:

$P(x = k) = \frac{\binom{M}{k}\binom{N-M}{n-k}}{\binom{N}{n}} $

Where N is our population (all balls), M are the sample we are interested, and N-M the others kind of balls collors in our box.

Balls are drawn until the three red balls are drawn and will happen 9 drawn. So we are intersting in the probability to find all the three balls with minimum trials three and max trials nine.

Doing this to three trials we obtain:

$P(x = 3) = \frac{\binom{3}{3}\binom{10}{0}}{\binom{13}{3}} = 0.0034 $

Doing this to four trials we obtain:

$P(x = 4) = \frac{\binom{3}{3}\binom{10}{1}}{\binom{13}{4}} = 0.013$

Until the last trial:

$P(x = 9) = \frac{\binom{3}{3}\binom{10}{6}}{\binom{13}{9}} = 0.29$

If we sum all the equations  $ P(D \le 9) = \sum_{i=3}^{9}\frac{\binom{3}{3}\binom{10}{i-3}}{\binom{13}{i}}=0.734265734$.

Is correct your calculus, maybe the question if badly formulated or is missing something else.