In a room with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?

1)5/21
2)3/7
3)4/7
4)5/7
5)16/21


how can we solve this one with combinations?

I would love a solution to this problem, as well.

I'll say the solution....after having seen some answers

Honestly, I don't even know if this is correct.

First, I grouped the pairings in to four categories. a "4" stands for someone with one friend, and a "3" stands for someone with two.

So the possible groupings are 44, 43, 34, and 33.

But these groupings are not all equally likely to happen; here are the probabilties for each arrangement:

44 = (4/7 x 3/6) = 12/42

43 = (4/7 x 3/6) = 12/42

34 = (3/7 x 4/6) = 12/42

33 = (3/7 x 2/6) = 6/42

OK. Next, I figured out, for each arrangement, the probability that the people would be friends.

For 44: The first 4 has a 1/6 chance of being friends with the second one.

For 33: The first 3 has a 1/3 chance of being friends with the second one.

Here's where I think I'm on the right track, but I'm not sure...

For 34 and 43: The 4 has a 1/6 chance of being friends with the 3, and the 3 has a 1/3 chance of being friends with the 4. So, I took the average of 1/6 and 1/3, which is 1/4, and concluded that that any given 4 and 3 have a 1/4 chance of being friends.

Next, I multiplied the probability of each pairing occuring by the probability that those two people are friends:

44 = 12/42 x 1/6 = 2/42
34 = 12/42 x 1/4 = 3/42
43 = 12/42 x 1/4 = 3/42
33 = 6/42 x 1/3 = 2/42

If you sum the probabilities in the last column (2/42 + 3/42 + 3/42 + 2/42), you get a 10/42 chance that two randomly chosen people will be friends. 10/42 reduces to 5/21.

So, if there is a 5/21 chance that two people will be friends, there is a 16/21 chance they will not be friends.

My answer is 16/21.

yes, you're right. OA is 16/21. If it helps, I give you my reasoning:

(4/7)(5/6) + (3/7)(4/6) if you choose one of the 4 with one other friend, then you have a 5/6 chance of not picking their friend 2nd. If you choose one of the 3 with 2 friends, you have a 4/6 chance of not picking one of their friends second. Add them up.

20/42 + 12/42
32/42 = 16/21

please cite the author of this question - teachers cannot reply until the author has been cited (and, actually, we'll have to delete the entire thread if the author is not cited). Thanks!

I took it from another forum

This question came from an MGMAT CAT. This is my third time seeing this question.

Thank you! Moved to appropriate section.

Some good discussion above already on this one. Another approach is below. A warning first: be careful about pulling random questions from web sites that don't have users cite the sources. If you are also using our tests, you've now exposed yourself to possible test questions ahead of time - do that enough, and the tests will lose validity for you!

Begin by counting the number of relationships that exist among the 7 individuals whom we will call A, B, C, D, E, F, and G.

First consider the relationships of individual A: AB, AC, AD, AE, AF, AG = 6 total. Then consider the relationships of individual B without counting the relationship AB that was already counted before: BC, BD, BE, BF, BG = 5 total. Continuing this pattern, we can see that C will add an additional 4 relationships, D will add an additional 3 relationships, E will add an additional 2 relationships, and F will add 1 additional relationship. Thus, there are a total of 6 + 5 + 4 + 3 + 2 + 1 = 21 total relationships between the 7 individuals.

You can also use the anagram grid or "real math" to get 7! / (5!2!) = 21.

We are told that 4 people have exactly 1 friend. This would account for 2 "friendship" relationships (e.g. AB and CD). We are also told that 3 people have exactly 2 friends. This would account for another 3 "friendship" relationships (e.g. EF, EG, and FG). Thus, there are 5 total "friendship" relationships in the group.

The probability that any 2 individuals in the group are friends is 5/21. The probability that any 2 individuals in the group are NOT friends = 1 – 5/21 = 16/21.