Calculating the probability of dying due to failed death saving throws in D&D 5e

Nat 20 = Auto-survive, so will not show up as a possible result (because I'm calculating chance of dying). .05 probability of occurrence

P = result between 10 and 19, inclusive, results in a Pass. .5 probability of occurrence

F = result between 2 and 9, inclusive, results in a Failure. .4 probability of occurrence

X = result of 1, results in 2 Failures. .05 probability of occurrence.

X, (F or X) = .05*.45 = .0225 chance +

X, P, (F or X) = .05*.5*.45 = .01125 chance +

X, P, P, (F or X) = .05*.5^2 *.45 = .005625 chance +

F, X = .4*.05 = .02 chance +

F, F, (F or X) = .4^2 *.45 = .072 chance +

F, F, P, (F or X) = .4^2 *.5*.45 = .036 chance +

F, F, P, P, (F or X) = .4^2 *.5^2 *.45 = .018 chance +

F, P, X = .4*.5*.05 = .01 chance +

F, P, F, (F or X) = .4^2 *.5*.45 = .036 chance +

F, P, F, P, (F or X) = .4^2 *.5^2 *.45 = .018 chance +

F, P, P, X = .4*.5^2 *.05 = .005 chance +

F, P, P, F, (F or X) = .4^2 *.5^2 *.45 = .018 chance +

P, X, (F or X) = .5*.05*.45 = .01125 chance +

P, X, P, (F or X) = .5^2 *.05*.45 = .005625 chance +

P, F, X = .5*.4*.05 = .01 chance +

P, F, F, (F or X) = .5*.4^2 *.45 = .036 chance +

P, F, F, P, (F or X) = .5^2 *.4^2 *.45 = .018 chance +

P, F, P, X = .5^2 *.4*.05 = .005 chance +

P, F, P, F, (F or X) = .5^2 *.4^2 *.45 = .018 chance +

P, P, X, (F or X) = .5^2 * .05*.45 = .005625 chance +

P, P, F, X = .5^2 *.4*.05 = .005 chance +

P, P, F, F, (F or X) = .5^2 *.4^2 *.45 = .018 chance =

.404875 total chance of dying, assuming no outside influences, or 40.4875% chance of death.