What is the probability that the first $5$ balls are blue? 
Four red, $8$ blue, and $5$ green balls are randomly
arranged in a line.
(a) What is the probability that the first $5$ balls are
blue?

Here is my solution,
Since there are $17$ balls we can randomly arranged in a line, Number of ways to arrange ($m$) $\dfrac{17!}{8!5!4!}$
Since first five balls are blue number of ways($n$) $\dfrac{12!}{4!3!5!}$
required probability is $\dfrac{n}{m}$
Is my answer correct?
 A: This is a hypergeometric distribution with parameters 8, 5, 5. The probability that X=5 can be found via (8 choose 5) * ( 9 choose 0) / (17 choose 5).
A: Direct calculation: $\frac{8\times 7\times 6\times 5\times 4}{17\times 16\times 15\times 14\times 13}=0.009049773755656$
Fraction $=\frac{2}{221}$.
A: I don’t understand your answer, most probably due to my lack of knowledge in probabilities.
However, i do understand that much:
Arranging the balls in a line is equivalent with attempting to draw balls from a bag, while discarding the ball just drawn.

*

*The odds of blue at first attempt are 8 in 17

*At second attempt are 7 in 16

*At third attempt are 6 in 15

*At fourth attempt are 5 in 14

*At fifth attempt are 4 in 13

The odds of making 5 consecutive attempts resulting in blue ball drawn are the result of the product of the previous individual attempts. This answer was already presented by someone else so I concur with it. I will not provide the formula, but I will provide the final answer since this is what you asked for: $\frac{2}{221}$
A: Your answer is quite correct. $$\cfrac{~\cfrac{12!}{3!5!4!}~}{\cfrac{17!}{8!5!4!}}=\dfrac{8!/3!}{17!/12!}$$
This is indeed the probability for obtaining $\mathit 5$ from $\mathit 8$ blue balls when selecting any $\mathit 5$ from all $\mathit 17$ balls to place in the first five positions (I.E. that blue is the colour of all balls in the first five positions).
$$\begin{align}\binom {8}5\div\binom{17}5 &= \dfrac{~~8\cdot~~7\cdot~~6\cdot~~5\cdot~~4}{17\cdot 16\cdot 15\cdot14\cdot 13}\\[1ex]&=\dfrac{2}{17\cdot 13}\\[1ex]&=\dfrac{2}{221}\end{align}$$
