A simple multiplicative approach is only valid if you require failures of all three redundant items to constitute a system failure.

if you need a 2 out of 3 multiple vote; it's a much higher probability than that because you're only allowed one failure. For an individual element reliability, R =(1-Pf), and all three elements being identical, the resulting 2 of 3 redundant reliability is given by

Rs = R^3 + 3.R^2.(1-R)

For P =0.1, Rs = 0.972, so Ps = 0.028, 28 times higher than Pf^3

And the smaller the Pf, the larger this ratio becomes.

I fear we're getting off the navigation track here, though...