Would someone be so kind as to vet the solution I've come up with for part a?

P(fail)=0.95

Let even A = at least 4 fails before a success

P(A)=1-P(at least 1 success occurs in the first four trials) = 1-P(B)

P(B) = P(at least one success occurs in the 1st 4 trials) = 1-P(no successes in the first four trials) = 1-(0.95)^4 = 0.18549

so P(A)= 1 - 0.18549

= 0.81451