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?
Best Answer
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).
$$\begin{align}\binom {8}5\div\binom{17}5 &= \dfrac{~~8\cdot~~7\cdot~~6\cdot~~5\cdot~~4}{17\cdot 16\cdot 15\cdot14\cdot 13}\\[1ex]&=\dfrac{2}{17\cdot 13}\\[1ex]&=\dfrac{2}{221}\end{align}$$