# 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 • if the first 9 draws include all of the red balls, then it would work – Xiangyu Chen Feb 13 '18 at 19:57
• Draw $9$ balls without replacement. So do not stop drawing earlier. Then $\{D\leq9\}$ is exactly the event that among them there are $3$ red balls. Not more than $9$ draws were necessary for that. – drhab Feb 13 '18 at 20:25

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 $$\frac15\cdot \frac{\binom{3}{2}\binom{10}{6}}{\binom{13}{8}} = \frac{\binom{3}{3}\binom{10}{9-3}}{\binom{13}{9}} - \frac{\binom{3}{3}\binom{10}{8-3}}{\binom{13}{8}}.$$

• That means we could use the method in the 2nd question of $P(D = 9)$ to find $P(D = 3), P(D = 4), ..., P(D = 9)$ and summing them to find $P(D \le 9)$. I think that logic is easier to understand. – A_for_ Abacus Feb 19 '18 at 3:36
• Personally I find the direct calculation of $P(D\leq9)$ much simpler than trying to prove that $P(D=k) = \frac{1}{14-k}\cdot\frac{\binom32\binom{10}{k-3}}{\binom{13}{k-1}} = \frac{\binom33\binom{10}{k-3}}{\binom{13}{k}} - \frac{\binom33\binom{10}{k-4}}{\binom{13}{k-1}},$ but that may be a matter of personal taste. (It may also be a conceptual issue, but that's one reason to study math--it broadens one's ability to conceptualize things in different ways. But if you're comfortable at least with the book's answer for part (a) now, I'm happy.) – David K Feb 19 '18 at 4:23

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.

• But we're not trying to find $P(D=9)$, we're trying to find $P(D \le 9)$. So there are cases where you can't choose 6 other balls. – A_for_ Abacus Feb 13 '18 at 20:10
• @A_for_Abacus yes you can. Just go on with drawing till you have $9$. Nobody will stop you. – drhab Feb 13 '18 at 20:34
• @A_for_Abacus drhab nailed it. At the end of the day, you've won if you have three reds out of nine. If you got the three reds by the sixth draw, well, you get a gold star, but it's still covered by the counting argument. If you drew the three reds with the first three draws, you get a gold star and a doggie sticker ... but it's still covered by the counting argument! Another way of putting it: Drawing three reds in no more than nine draws is different than drawing the third red on the ninth draw, which is what you may be thinking? – John Feb 13 '18 at 20:50
• Okay, but we should be able to calculate $P(D \le 9)$ by summing $P(D = 3) + P(D=4) +...+P(D=9)$, what would the formula for those values be in this case? – A_for_ Abacus Feb 14 '18 at 0:13
• @A_for_Abacus As in the answer to part (b), $P(D=k) = P(D\leq k) - P(D\leq k-1) = \binom{10}{k-3}/\binom{13}{k} - \binom{10}{k-4}/\binom{13}{k-1}$ for $k>3,$ and $P(D=3) = P(D\leq 3) = 1/\binom{13}{3}.$ – David K Feb 14 '18 at 1:02

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.

• What you calculate for $P(x=4)$ is actually $P(x\leq 4).$ – David K Feb 14 '18 at 1:04