Candies withdrawal probability for a particular subsequence You are taking out candies one by one from a jar that has 10 red candies, 20 blue candies, and 30 green candies in it. What is the probability that there are at least 1 blue candy and 1 green candy left in the jar when you have taken out all the red candies?
 A: I came across following solution but curious to see how others approach this problem:
Let $T_r$, $T_b$ and $T_g$ be the "draw number" of last red, blue and green candies drawn respectively. So the required probability here is $P(T_r<T_b \cap T_r< T_g)$
Now the event $T_r<T_b \cap T_r< T_g$ can be achieved in two different and independent ways, i.e., $T_r < T_b < T_g$ and $T_r < T_g < T_b$
$\therefore P(T_r<T_b \cap T_r< T_g) = P(T_r < T_b < T_g) + P(T_r < T_g < T_b) = \frac{20}{30} * \frac{30}{60} + \frac{30}{40} * \frac{20}{60} = \frac{7}{12}$
A: Indeed.   We consider what would happen were we not to stop after drawing the last red candy.
The probability that the last of sixty candies drawn is one of the thirty green candies is $\tfrac{30}{60}$.   Ignoring the green candies, the probability that the last of the thirty other candies drawn is one of the twenty blue is $\tfrac{20}{30}$.
Likewise the probability that the last draw would be blue and the last not blue draw would be green is $\tfrac{20}{60}\tfrac{30}{40}$ .
Thus the probability that you seek is clearly: $\tfrac{30}{60}\tfrac{20}{30}+\tfrac{20}{60}\tfrac{30}{40} =\tfrac 7 {12}$
A: The following is almost certainly overkill, but what the heck.
If you think of picking red, blue, and green as corresponding to taking steps in the $x$, $y$, and $z$ directions on a three-dimensional lattice, each way of drawing candies corresponds to a path from $(0,0,0)$ to $(10,20,30)$ of steps that go one unit in a positive direction each.
In the picture below, a candy-sequence would correspond to a path of adjacent cubes from bottom left to top right that moves right, up, or away from the viewer.

If you reach the right-most face in the red area, you’ve run out of red candies first, so to answer your question, you want to know how many paths (going only right, up, or away) there are from the origin to the upper-right box that reach the red area.
Alternatively, you would want to know what fraction of the $\frac{60!}{10!\cdot20!\cdot30!}$ sequences do not correspond to paths that lie within the following figure.

You can often write recurrence relations for counting problems like this, but here it’s far more work then needed, and you seem to have found a simple solution.
A: $P(T_r<T_b)$ is not that simple to calculate. 
So I was counting paths. Basically for given number of $r,b,g$, we have $\frac{(r+b+g)!}{r!\cdot b!\cdot g!}$ paths. Denote number of all path where $T_r<T_b$ as $F(r,b,g)$, we have: 
$$
F(r,b,g)=P_g(T_r<T_b)\frac{(r+b+g)!}{r!\cdot b!\cdot g!} 
$$
and
$$
F(r,b,g)=F(r-1,b,g)+F(r,b-1,g)+F(r,b,g-1)
$$
We can easily verify that $P_g(T_r<T_b)\neq \frac{b}{r+b}$
