In a room filled with 7 people, 4 people have exactly 1

[JonathanSchneider]

Nice work, Matt. I like your creativity on this one. Not the easiest approach, probably, but very cool regardless.

[chetna.response]

Hi,

Can this question be answered using the following approach?

Let the groups of friends be:
A-B (group-1)
C-D (group-2)
E-F-G (group-3)
As seen, this satisfies the given conditions in the question.

Now the possible scenarios which satisfy the condition that two individuals picked up are random are NOT friends are:

(one person picked from group-1 AND one person picked from group-2)
OR (one person picked from group-2 AND one person picked from group-3)
OR (one person picked from group-3 AND one person picked from group-1)

This implies, the probability that two individuals picked up are random are NOT friends is:

[(2C1 * 2C1) + (2C1 * 3C1) + (3C1 * 2C1)] / 7C2
= [(2*2) + (2*3) + (3*2)] / 21
= [4+6+6]/21 = 16/21

Is this approach correct? Please can anyone confirm this?

Thanks a lot.

[jlucero]

It sure is. Notice that there are 4 combinations of g1/g2: AC, AD, BC, BD, 4 of g1/g3: AE, AF, AG, BE, BF, BG, 3 of g2/g3: CE, CF, CG, DE, DF, DG. And 21 pairs overall: 7c2. Good stuff.

[chetna.response]

thanks!

[tim]

:)