4
$\begingroup$

This same problem also appears here: http://perplexus.info/show.php?pid=9271

A drawer contains a mixture of red socks and blue socks, at most $2014$ in total. It so happens that, when two socks are selected randomly without replacement, there is a probability of precisely $\frac23$ that both are red or both are blue.

What is the maximum possible number of red socks in the drawer that is consistent with this data?

$\endgroup$
  • $\begingroup$ When you say precisely, that is indeed an exact value? And 2/3 is the chance of (Blue/Blue OR Red/Red)? $\endgroup$ – oerkelens Nov 7 '14 at 10:47
  • $\begingroup$ @oerkelens, here is a possible solution as an example: You could have 20 red socks and 5 blue socks. The prob of both red = 20*19/25*24 plus prob of both blue = 5*4/25*24 = 2/3. $\endgroup$ – Kenshin Nov 7 '14 at 10:49
5
$\begingroup$

Answer is:

The answer is $1065$ red and $285$ blue socks.
Probability of getting same colours = $\frac{1065 \times 1064 + 285 \times 284}{1350 \times 1349} = \frac23$

Inequation and equations are

$r + b \le 2014$
$r \ge 0, b \ge 0$
$3 \times (r^2 - r + b^2 - b) = 2(n^2 - n)$
Solving all, $r = 1065$, $b = 285$ and $n = 1350$

$\endgroup$

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.