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A box contains $20$ balls numbered $1,2,3,...,20$. If 3 balls are randomly taken from the box, without replacement, what is probability that one of them is the average of the other two?

Which one of these is the correct answer?

  • 7/45
  • 3/38
  • 6/155
  • 3/154
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    $\begingroup$ What do you think the answer is, and why? This forum is very much about "show your attempt" such that others - very clever people I might add, I have learned much - can review what you have done, and suggest improvements. $\endgroup$
    – Russ
    Nov 14, 2018 at 12:54
  • $\begingroup$ I couldn't agree more, this is a problem which I couldn't solve by myself. I thought that it'll be just time wasting for people trying to help, to see pointless attempts of someone, that don't even closely lead to the solution. :( $\endgroup$
    – Sth
    Nov 14, 2018 at 13:06
  • $\begingroup$ I understand your hesitance, but this board is fairly conservative, and they are inundated with "do my homework for me" kinds of requests. No matter how poor you think your attempt is, at least it shows that you've tried, separating you from academically immature posters. $\endgroup$
    – Russ
    Nov 14, 2018 at 14:19

3 Answers 3

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We need to determine the number $N$ of 3-tuples from this set of numbers that have the desired property that one number is the average of the other two (We also need the total number of 3-tuples from this set, but think isn't a challenging task).

Let's first think about what is needed for a number $n$ to be the average of two other numbers $a$, and $b$. Based on the definition of average, we have $$n=\frac{a+b}{2}$$

A way to look at this equation is to say that $n$ is the midpoint of the interval $[a,b]$. Therefore, to find $N$ all we need to do is figure out how many intervals are possible for each number in our set, and then add them all up.

First, we don't need to consider $1$ or $20$, because they do not have an interval around them for this set. For the other numbers, the number of intervals just depends on the closest distance that number is from either $1$ or $20$. More specifically, we need the smaller distance of the two.

Therefore, $$N=\sum_{n=2}^{19}\text{Min}(20-n,n-1)$$

But we can easily determine which one is the minimum for a given number. When $n\leq10$, $n-1$ is the minimum, and when $n\geq11$, $20-n$ is the minimum.

Therefore, $$N=\sum_{n=2}^{10}(n-1)+\sum_{n=11}^{19}\text(20-n)$$

These sums should be easily manageable now. I will leave the rest of the work to you.

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The number of 3 long arithmetic sequences is:

Let $a_d$ be a number of such sequences if difference is $d$

$d=1:\;\;\;$$(1,2,3);(2,3,4),...(18,19,20)$, so $a_1=18$.

$d=2:\;\;\;$$(1,3,5);(2,4,6),...(16,18,20)$, so $a_2=16$.

$d=3:\;\;\;$$(1,4,7);(2,5,8),...(14,17,20)$, so $a_3=14$.

$\dots$

$d=9:\;\;\;$$(1,10,19);(2,11,20)$, so $a_9=2$.

So the number of all good sequences is $2+4+...+18 = 90$. Thus the answer is $$P= {90\over {20\choose 3}} = {90\over 20\cdot 19\cdot 3}= {3\over 38}$$

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Refer to the table: $$\begin{array}{c|c|c} \text{Middle (average) number}&\text{Sets}&\text{Number of sets}\\ \hline 2&\{1,2,3\}&1\\ 3&\{1,3,5\},\{2,3,4\}&2\\ 4&\{1,4,7\},\{2,4,6\},\{3,4,5\}&3\\ \vdots&\vdots&\vdots\\ 9&\{1,9,17\},\{2,9,16\},\cdots,\{8,9,10\}&8\\ 10&\{1,10,19\},\{2,10,18\},\cdots,\{9,10,11\}&9\\ 11&\{2,11,20\},\{3,11,19\},\cdots,\{10,11,12\}&9\\ 12&\{4,12,20\},\{5,12,19\},\cdots,\{11,12,13\}&8\\ \vdots&\vdots&\vdots\\ 18&\{16,18,20\},\{17,18,19\}&2\\ 19&\{18,19,20\}&1\\ \hline \text{Total}&&2\cdot \frac{(1+9)\cdot 9}{2}=90 \end{array}$$ Hence, the required probability is: $$\frac{90}{{20\choose 3}}=\frac{90\cdot 1\cdot 2\cdot 3}{20\cdot 19\cdot 18}=\frac{3}{38}.$$

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