<PRE><B>Question 1134100</B>
15 total
for four mathematicians only
8/15*7/14*6/13*5/12=0.0513

equal would be 2 of each
total number of ways to choose any four is 15C4=1365
for the numerator 8C2*7C2=28*21=588
the probability is 588/1365=0.4307
At least two mathematicians 
for 3 would be 8C3*7C1=56*7=392, and that probability is 0.2872
for 4, 0.0513
So at least two would be the sum of three terms or 0.7692</PRE>