For Example 13 in Chapter 14 (pg. 412-3), Peter Visscher (University of Edinburgh) has 
correctly pointed out that that we need not assume equal probabilities for each of the two
events.  If we do not assume this, one can obtain a slightly lower total n.

In Peter's words: 

 one could minimize the complete function

f(n1 + n2)= [ 1 - (1-2c)^n1][ 1 - (1-c)^n2] > P

For c=0.01, P=0.95, I get {from a quick search} n1=196, n2=344,
N=540 (only slightly smaller than your 555).