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

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!

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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

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

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

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).

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Awesome analysis and intresting part about singaporeans :o –  Yellow Skies Feb 26 '13 at 9:30