1
$\begingroup$

The question is

What is the probability that a subset of $\{1,2,\dots,12\}$ of size 6 contains the number 1?

The brute force method is to count how many subsets 1, which amounts to counting how many subsets of size 5 there are in $\{2,\dots,12\}$, i.e. $\binom{11}{5}$. Hence the answer is $$\frac{\binom{11}{5}}{\binom{12}{6}} = \frac12.$$ But how could one see that there is an immediate bijection between subsets containing 1 and not containing 1?

$\endgroup$
3
$\begingroup$

The bijection you want is just: "map a subset to its complement". Easy to see that, given a subset $S$ and its complement $\overline S$, then $1$ is in exactly one of the two.

$\endgroup$
1
$\begingroup$

One insight is that selecting $6$ out of $12$ items gives probability one-half that the particular number $1$ falls in the selected subset (because selected and leftover subsets are equal in size).

$\endgroup$
  • $\begingroup$ does this mean that if the subsets were of size 4, there would be a $1/3$ chance of selecting 1? $\endgroup$ – user369210 Aug 30 '16 at 21:07
  • $\begingroup$ Yes, that is a correct generalization. $\endgroup$ – hardmath Aug 30 '16 at 21:08
  • 1
    $\begingroup$ ha! This is such a stupidly easy question I have no idea why I overcomplicated it $\endgroup$ – user369210 Aug 30 '16 at 21:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.