This was tough.  There were many incorrect solutions submitted.  The correct probability is about 41.89%.

First compute the number of ways in which both members of Team 1 pair up in the same lane:  There are 2ways to choose a member of Team 1, 6 lanes into which that runner can go, 1 way to place the second runner from that school, and 10! ways to distribute the remaining runners into the remaining lanes for a total of 2·6·10! ways.  The probability that Team 1 will have both of its runners in a lane therefore 2·6·5!/10! = 1/11.  

Next compute the number of ways in which both members of Team 1 and Team 2 will pair up.  Repeating the logic of the previous sentence, that number will be 2·6·2·5·8!, giving a probability of 1/(11·9) of this even occurring.  

In general, the probability P(k) that k teams will have both of its members pair up is given in the table on the left above.  Using the inclusion-exclusion principle you can then compute the desired probability that at least one of the teams will pair up is 

= .4189... 

You are visitor number 4005 to this page.
ã2003 Alberto L. Delgado