Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A park contains 4 loving couples. 3 people are randomly selected. The chance of picking a couple amongst the 3 is...

My solution

1st Move: Select any person 8 ways.

2nd Move: Select the person’s mate 1 way

3rd Move: Select any person 6 ways.

Total Ways 8*6*(3!/2!) ----- where (3!/2!) is the permutation factor.

Sample Space 8P3

Therefore, Probability is = 0.429

However, it seems obvious that my probability cannot be this high. Did I make a mistake somewhere? Thanks stack!

share|cite|improve this question
I arrived at the same answer differently by considering the complementary event of no lovey-doveys getting included. Then there are 8 alternatives for the first person, 6 for the second and 4 for the last as opposed to 8,7,6 in the unconstrained case. $$P=1-\frac{8\cdot6\cdot4}{8\cdot7\cdot6}=\frac37\approx0.4286$$ – Jyrki Lahtonen Feb 26 '13 at 8:59
Thanks! Make it an answer so i can drop a few point later! – Yellow Skies Feb 26 '13 at 9:05
However, do you think the answer is likely to be as high as 0.4 – Yellow Skies Feb 26 '13 at 9:07
up vote 1 down vote accepted

Here is my solution. Consider the number of ways you can select 3 people out of a set of 8.

The total number (in your terms, I believe, is unrestricted) of combinations is:

$$ ^8C_3 $$

Consider the number of ways to select a set of 3 people with 1 couple. Every couple selected can have 1 out of the remaining 6 people. There are 4 couples. Hence, the number of combinations is

$$4 \times 6$$

Hence the resultant probability is

$$ \frac{4\times6}{{8\choose3}} = 0.429 $$

I think your answer is correct.

Being Singaporean, I believe you still have doubts, based on my interaction with Singaporean students. Let me double-confirm (if it's the right way of saying) your answer.

Let's now consider that of the 4 couples, I select any 3 couples. The number of ways is $$^4C_3 = 4$$

Of a set of 3 couples, you can select 1 man or 1 woman each to form a set of 3 people who are not couples. The total number of ways is

$$2^3 = 8$$

Thus, the number of ways to select 3 people who are not couples is:

$$4 \times 8 = 32$$

If you add $32$ and $24$, you get $^8C_3 = 56$, which is the total number of combinations. The answer is indeed correct since the set of 3 either contains a couple or does not (they, combined form the universal set for this question).

share|cite|improve this answer
Awesome analysis and intresting part about singaporeans :o – Yellow Skies Feb 26 '13 at 9:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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