Probability game with 17 balls

Question

We place randomly 17 balls in a row. 10 blue, 6 red and 1 yellow. Suppose there places are numbered 1 to 17 where 1 is the first place on the left.

What is the probability that the yellow ball will have at most 3 red balls before him.

I tried to break it into different events. Let $\ A_i $, $\ i = 0,1,2,3 $ be the event that there is $\ i $ red balls left to the yellow ball. The events are mutually exclusive so $\ P(A_1 \cup A_2 \cup A_3 \cup A_4) = P(A_1) + P(A_2)+ P(A_3) + P(A_4) $

So the probability that there are $\ 0 $ red balls left to the yellow ball is determined by picking 7 spots, then placing the red ball on most left spot and then placing the red balls in the spots that left and then the blue balls : $$\ P(A_0) = \frac{{17 \choose 7} \cdot 1 \cdot 6! \cdot 10!}{17!} = \frac{1}{7} $$

and the event of having one red ball before the yellow should be:

$$\\ P(A_1) = \frac{{17\choose 7}\cdot 1 \cdot 1 \cdot 5! \cdot 10!}{17!}$$

But apparently I'm wrong about this one as the answer is $\ \frac{4}{7} $

I don't understand what am I missing?

Answer 1

Among the $7$ red and yellow balls the yellow ball can have any rank with equal probability. The probability that its rank is $\leq4$ therefore comes to ${4\over7}$.

Answer 2

First you choose 10 places to put blue balls. That you can do on $${17\choose 10}$$ ways. Now you put $k$ red balls, where $k\in\{0,1,2,3\}$ on first $k$ free places (which are already $\color{blue}{\rm uniqely\; determined}$ by blue balls) and then yellow and then the rest of balls. So you have $${17\choose 10} \cdot 4$$ good configurations among all $${17\choose 10} {7\choose 6} $$

configurations. So the result is $${4\over 7}$$