Two friends compete with each other and five other, equally good, violinists for first and second chair in an orchestra, in a blind competition What is the probability that the two friends end up as first and second chair together?

Assuming all violinists have an equal chance of getting any chair, we can calculate the probability of the two friends getting the first and second chair together among the seven violinists.

The total number of ways to choose the first and second chair violinists from the seven people is a permutation because the order matters: first we choose someone for the first chair, and then someone different for the second chair. The number of ways to do this is 7 choices for the first chair times 6 remaining choices for the second chair, which is 7 * 6 = 42 possible combinations.

However, we are only concerned with the combinations where the two friends are chosen. If we consider one particular friend for the first chair, there's only one other choice for the second chair, the other friend. The same is true if we choose the other friend first. So there are 2 scenarios where the friends get the first and second chair (first friend then second friend, or second friend then first friend).

So the probability that the two friends end up as first and second chair together is:

Number of favorable outcomes (both friends in the first two chairs) / Total number of possible outcomes (any two people in the first two chairs) = 2 / 42 = 1 / 21

Therefore, the probability is 1/21.