# Probability game with 17 balls

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?

• You can safely ignore the blue balls in this question. If you place 6 red balls and 1 yellow ball in some random order in a line, what is the probability that the Yellow ball in the first 4 spots of the line? – Doug M Oct 29 '18 at 17:50

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 $${4\over7}$$.
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 $${4\over 7}$$