10 red balls, 20 yellow balls, 30 blue balls in a box, what is the probability of first empty all the red balls out?

Question

Suppose take one ball out at a time, and we do not put it back. What is the probability that all the red balls are taken out while there are both yellow and blue balls left.

Clarification exmaple, at one particular time, we picked a red ball, now the status is that:

out of box: 10 red balls, m yellow, n bule. where 20>m >=0, 30>n>=0
in the box: 0 red balls, 20-m yellow, 30-n blue

we say that we have all the red balls taken out while there are some blue and yellow balls in the box.

I think we can do like this, we have $10$ red balls, $m<20$ yellow balls, and $n<30$ blue balls out of the box. The total balls is $10+m+n$, the probability of getting into this situation is:

$$ P_{m,n}=\frac{\binom{10}{10}\binom{20}{m}\binom{30}{n}}{\binom{60}{10+m+n}} $$

The total probability therefore is:

$$ P=\sum_{m=0}^{19}\sum_{n=0}^{29} P_{m,n} $$

Is my proposal right? seems haven't used the "not put it back condition..."

How to sum this? Is there any simpler method?

Answer 1

Hint:

Split the yellows up in $10$ light-yellows and $10$ dark-yellows.

Split the blues up in $10$ light-blues, $10$ middle-blues and $10$ dark-blues.

Let e.g. $E_{(y_l,r,b_d,b_m,y_d,b_l)}$ denote the event that first the light-yellows are taken out, then the red, then... et cetera.

There are $6!=720$ of that sort of events. They are disjoint, covering and equiprobable.

Now find out how many of these events are such that - if the reds are taken out - there are still yellows and blues in the box.

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edit to make things more clear.

Let $E$ denote the event that - if the reds are taken out - there are still yellows and blues in the box. Then you are looking for $\Pr(E)$. This $E$ can be written as a union of the disjoint events described above. Now let's see how many of them belong to this union:

$5!=120$ events of the form $E_{(r,-,-,-,-,-)}$.

$5!=120$ events of the form $E_{(-,r,-,-,-,-)}$.

$5!-2!3!=108$ events of the form $E_{(-,-,r,-,-,-)}$.

$72$ events of the form $E_{(-,-,-,r,-,-)}$.

No events of the form $E_{(-,-,-,-,r,-)}$.

No events of the form $E_{(-,-,-,-,-,r)}$.

So we end up with probability: $$\Pr(E)=\frac{420}{6!}=\frac7{12}$$

Answer 2

I would say

$P=\frac{10}{60}\cdot\frac{9}{59}\cdot\frac{8}{58}\cdots\frac{3}{53}\cdot\frac{2}{52}\cdot\frac{1}{51}=\Pi_{i=0}^{9} \frac{10-i}{60-i}\doteq 1.33\cdot 10^{-11}$

Fixed a number of balls in total (=60)

P = the likelihood of dragging all the red balls in the first 10 moves.

Answer 3

We have $n=r+b+y$ balls (with hidden numbers on them) in a random order, hence a sample space of $n!$ equiprobable events. An event produces a word over the alphabet $\{R,B,Y\}$, and is favorable if this word begins with $B^jY$ for some $j\in[b]$, or with $Y^k B$ for some $k\in[y]$.

Words beginning with $B^jY$ can be formed in $b(b-1)\cdots\bigl(b-(j-1)\bigr)y(n-j-1)!$ ways: We first arrange in turn $j$ selected blue balls, then a selected yellow ball, and finally arrange the remaining balls in an arbitrary way. Similarly for the words beginning with $Y^k B$.

It follows that the total number of favorable words is $$\sum_{j=1}^b y {b\choose j}j!(n-j-1)!+\sum_{k=1}^y b {y\choose k}k!(n-k-1)!=:N\ ,$$ and the probability $p$ in question then computes to $$p={N\over n!}\ .$$

Answer 4

Do the case with $2$ colors first: If there are $a$ red balls and $b$ yellow balls, the probability that red is out first is $b/(a+b)$ (the probability that the last ball is yellow).

If there are $a$ red, $b$ yellow and $c$ blue, distinguish cases according to the last ball, and reduce to the case of $2$ colors:

• If the last ball is yellow (probability $b/(a+b+c)$) then imagine conditioning on all the possible arrangements of the yellow balls. The red and blue balls need to go in the gaps, and the probability that red is out first is $c/(a+c)$ (independent of the arrangement of the yellow balls).

• If the last ball is blue (probability $c/(a+b+c)$) the probability that red is out first, is similarly $b/(a+b)$.

So the answer is $$\frac{b}{a+b+c} \cdot \frac c{a+c} + \frac{c}{a+b+c} \cdot \frac b{a+b} \,,$$ which works out to be $\frac7{12}$ in this case.