It's impossible to know – one person has 70 friends but we have no way of knowing how likely any of them could also be friends of each other. You need to introduce some sort of way that these people interact that we are aware of. Also, the fact that there are 100 is irrelevant because you're only asking about 1. Friendship is never random btw!  :) 

Yes, Andrew. You are right. I think that is why I can't find the answer. I wanted to make sure that there is actually no answer for this problem.

I'm pleased I got the right answer Sarah lol :)

Thanks Andrew. I hope we are both right. Lets see if someone finds another answer. lol

Yes lol. I'm in New Zealand btw. It's a very easy life here and beautiful :)

looking at this another way.  The probability that any one other person is one of his friends is 70/99 (I assume you don't count being friends with himself!) but this is true of any other person, so the probability of any two of his firends being friends is also 70/99. The probabiity that there will be at least one pair of his friends that are also friends is 1 minus the probability that all pairs of his friends (70x69/2= 2415) are not friends so is 1 minus (29/99)to the power 2415 - practically I. (This is the classic birthday problem in a different context.)

 The solution presented by Hugh is really interesting. However I want to solve it in a different manner with some minor corrections.

According to the question any person (like A) has 70 friends. We call two of them B and C. As Hugh said the probability that any two persons be friend is 70/99 but for B and C removing their common friend (A) this probability reduces to 69/98.

If we are talking about specific B and C then the probability is 69/98.

If the question is about the chance of friendship between any B and C( with A as a common friend) we require more analyses. Defining success as “B and C are friends” then the probability of success is p=69/98=0.70408. Then the total number of events (here relationships) is n=69*68/2=2346. We are seeking the probability of success between these events.

Mathematically in this problem, the Binomial distribution can be used. According to this distribution the probability of m success between n events when each event has the success probability of p is

P(n,m,p) = (n@m) * p^m *(1-p)^(n-m)

Where (n@m) is  binomial coefficient.

In our case we should calculate

Prob =  P(2346,1,0.70408) + P(2346,2,0.70408) + … + P(2346,2346,0.70408)

Or equivalently

Prob = 1 – P(2346,0,0.70408)

Hi

What is meant by "uniformly random friends?"

I know about uniform random numbers, but I it is not clear whether or not the expression "uniformly random friends" defines a probability distribution!

Yes, Mohamed. I meant uniform distribution.