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?
Best Answer
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.
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: $$\Pr(E)=\frac{420}{6!}=\frac7{12}$$