A group of $2n$ boys is to be divided into two groups of $n$ boys . What is the probability that the two tallest boys are in different groups ?

This is how I attempted it:

The probability that the two boys are in same group can be obtained as follows:

First we separate those two particular boys leaving us with $2n-2$ boys. We then form a group of $n$ boys not containing the two particular boys giving us a group $n-2$ boys in which the two boys can be accommodated. The probability of forming such groups is $\frac{\binom{2n-2}{n}}{\binom{2n}{n}}$. Thus the actual probability of forming the groups with the two boys in different groups is $1- \frac{\binom{2n-2}{n}}{\binom{2n}{n}}$ . However there seems a problem with this. Could you please point out where I was wrong ?

Also , please note that I already know the correct solution to this problem. I just wanted to correct my mistake.

Thanks for your help !

  • 1
    $\begingroup$ Now these two tallest could be in either group, is it not? $\endgroup$ – Satish Ramanathan Sep 30 '18 at 17:01
  • $\begingroup$ Yes but I thought that each group is identical right ? Because the probability of selecting each boy is equally likely , I guess $\endgroup$ – Aditi Sep 30 '18 at 17:02
  • $\begingroup$ Any group of $n-2$ boys that we form would have boys being equally likely to be selected $\endgroup$ – Aditi Sep 30 '18 at 17:04

Let us assume that the tallest boys are Andrew and Bruce. The configurations of this kind $$ (A\text{ together with }n-1\text{ other people })\quad (B\text{ together with }n-1\text{ other people }) $$ are $\binom{2n-2}{n-1}$ (it is enough to select Andrew's mates), while the configurations of this kind $$ (A,B\text{ together with }n-2\text{ other people })\quad (n\text{ other people }) $$ are $\binom{2n-2}{n-2}=\binom{2n-2}{n}$ (it is enough to select Andrew and Bruce's mates). The wanted probability is so

$$ \frac{\binom{2n-2}{n-1}}{\binom{2n-2}{n-1}+\binom{2n-2}{n}}=\frac{\binom{2n-2}{n-1}}{\binom{2n-1}{n-1}}=\color{red}{\frac{n}{2n-1}}. $$

  • $\begingroup$ Thank you very much :) now I understood ! $\endgroup$ – Aditi Sep 30 '18 at 17:25

Given that $2n$ boys are randomly divided into two subgroups containing $n$ boys each.

So, we can choose $n$ boys from $2n$ boys in $\dfrac{\dbinom{2n}{n}}{2!}=\dfrac{(2n)!}{2!\times(n!)^2}$ ways.

We need to find the probability of two tallest boys in different groups.

So, if we seperate $2$ tallest boys from $2n$ boys we have $2n-2$ boys.

Now we can arrange the group of boys in $\dfrac{\dbinom{2n-2}{n-1}}{2!}=\dfrac{(2n-2)!}{((n-1)!)^2\times 2!}$ ways

Now we can separate these $2$ tallest boys into $2$ different groups in $\dbinom{2}{1}$ ways.

So, in total we have $\dfrac{(2n-2)!}{((n-1)!)^2\times 2!}\times \dbinom{2}{1}$ ways

The required probability is $\dfrac{\dfrac{(2(n-1))!\times2}{((n-1))^2\times2!}}{\dfrac{(2n)!}{(n!)^2\times2!}}=\dfrac{n}{2n-1}$

  • $\begingroup$ Thank you for your help :) but could you please point out where I went wrong ? $\endgroup$ – Aditi Sep 30 '18 at 17:17
  • $\begingroup$ @Aditi I think you made a mistake in forming the groups after separating the two boys. $\endgroup$ – Key Flex Sep 30 '18 at 17:21
  • $\begingroup$ Ohh okay thank you ! $\endgroup$ – Aditi Sep 30 '18 at 17:26

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