No, that's correct.
Thanks so much for the screenshot. Much more artistic than a mere cut & paste. Who cares about usability after all?
If (1-p[I])^a
is 0.99999999999783207310676356492969
And probLessThanR
is 2.1679268932410045e-12
Then probLessThanR
is 0.0000000000021679268932410045
So using elementary school addition with carry:
(1-p[I])^a + probLessThanR
is, where
0.99999999999783207310676356492969
+ 0.0000000000021679268932410045`
~ 0.9999999999999999999999045684...
Carry 1 1
= 1.0000000000000000000000045694...
So 1-(1-p[I])^a - probLessThanR
is 0.0000000000000000000000045694...
. Which is what you have.