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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?

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  • $\begingroup$ What is the result of $\frac{n}{m}$? $\endgroup$ – Snoop Apr 25 at 20:46
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    $\begingroup$ Yes your working is correct. $\endgroup$ – Math Lover Apr 25 at 20:50
  • $\begingroup$ @MathLover slader.com/discussion/question/… $\endgroup$ – puka Apr 25 at 20:51
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    $\begingroup$ Yes it is the same thing. You could have considered them all different. Your working will give the same result. $\endgroup$ – Math Lover Apr 25 at 20:56
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    $\begingroup$ @Got it thank you very much $\endgroup$ – puka Apr 25 at 20:57
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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).
$$\begin{align}\binom {8}5\div\binom{17}5 &= \dfrac{~~8\cdot~~7\cdot~~6\cdot~~5\cdot~~4}{17\cdot 16\cdot 15\cdot14\cdot 13}\\[1ex]&=\dfrac{2}{17\cdot 13}\\[1ex]&=\dfrac{2}{221}\end{align}$$

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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).

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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.

  1. The odds of blue at first attempt are 8 in 17
  2. At second attempt are 7 in 16
  3. At third attempt are 6 in 15
  4. At fourth attempt are 5 in 14
  5. 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: $\frac{2}{221}$

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  • $\begingroup$ Quibble: The term "odds" in conjunction with the figures used is incorrect. GOK when this bad habit will be eradicated, if ever ! $\endgroup$ – true blue anil Apr 26 at 6:01
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Direct calculation: $\frac{8\times 7\times 6\times 5\times 4}{17\times 16\times 15\times 14\times 13}=0.009049773755656$

Fraction $=\frac{2}{221}$.

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