In the box are $7$ white socks, $5$ red socks and $3$ black socks.

$2$ socks are considered a pair if they have the same color.

$5$ arbitrary socks are selected at random from the box.

Find the probability that among the selected socks there are a white pair and a red pair.

I tried to solve this way:


I think that I have to choose $2$ out of $7$ white socks, then $2$ out of $5$ red socks, then $1$ out of the remaining $15-2-2$ socks. Then I need to divide it by the number of ways to choose $5$ out of $15$ socks.

But it is not correct.

  • 1
    $\begingroup$ Why did you choose that particular set of binomial coefficients as your solution? I assume you had some reasoning behind it, and didn't just pick random numbers out of a hat (or a box). If so, please explain that reasoning. $\endgroup$ – Ilmari Karonen Apr 26 '15 at 18:06
  • $\begingroup$ By socks in the first sentence, do you mean a pair of socks? English is not my mother tongue. $\endgroup$ – Haider Rehman Butt Apr 26 '15 at 18:08
  • $\begingroup$ @IlmariKaronen, I think that I have to choose two white socks from probable seven socks then two red from five socks and one any sock from all remaining (15-2-2) socks then I must divide it to all combinations of choice of five socks from 15 probable socks. $\endgroup$ – Timofey Apr 26 '15 at 18:22
  • $\begingroup$ @HaiderRehmanButt, In the first sentence there are not pairs, there are just single socks. $\endgroup$ – Timofey Apr 26 '15 at 18:22

The total number of ways to choose $5$ socks is $\binom{7+5+3}{5}=3003$.

The number of ways to choose $2W+2R+1B$ is $\binom72\cdot\binom52\cdot\binom31=630$.

The number of ways to choose $2W+3R+0B$ is $\binom72\cdot\binom53\cdot\binom30=210$.

The number of ways to choose $3W+2R+0B$ is $\binom73\cdot\binom52\cdot\binom30=350$.

Hence the probability of a white pair and a red pair is $\frac{630+210+350}{3003}\approx39.627\%$.

  • $\begingroup$ Thank you for right solution. I understood my fault. $\endgroup$ – Timofey Apr 26 '15 at 18:27
  • $\begingroup$ @Timofey: You're welcome :) $\endgroup$ – barak manos Apr 26 '15 at 18:32

Your answer is correct if the question is to be interpreted literally. It would include situations where the fifth sock is also perhaps red or white. If the questioner intended that there should be precisely two red socks and two white socks, with the third therefore being black, then replace the 11 in the numerator with 3.


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