A few people have said things that are similar while I was typing this up, but hopefully this still helps: I think the problem is that you’re implicitly assigning a probability of 0 to anything other than that one rate. In usual Bayesian analysis, you could imagine the launches as being analogous to a “biased coin.” Often success/failure scenarios are modeled as binomial, with a beta distribution describing our degrees of belief for what the “coin” will do. But for simplicity’s sake, let’s suppose we know that either 1% or 10% of the launches will fail, and we have no further information on what will happen, so we assign a probability of 0.5 to both rates of failure. Because either one or the other has to be the case (by the unrealistic setup of this problem), the unconditional probability of a failure is P(F) = 0.5(0.01) + 0.5(0.10) = 0.055
Now, we see a successful launch. Intuitively, this is more likely if the rate of failure is lower, so this should favor, for the rate R, the hypothesis R = 0.01 and decrease the probability that R=0.1:
P(R = 0.01|S) = P(S|R =0.01)P(R = 0.01)/P(S) = 0.99(0.5)/0.945 = 0.524
And since it has to be one or the other, P(R = 0.1|S) = 1 − 0.524 = 0.476
A few people have said things that are similar while I was typing this up, but hopefully this still helps: I think the problem is that you’re implicitly assigning a probability of 0 to anything other than that one rate. In usual Bayesian analysis, you could imagine the launches as being analogous to a “biased coin.” Often success/failure scenarios are modeled as binomial, with a beta distribution describing our degrees of belief for what the “coin” will do. But for simplicity’s sake, let’s suppose we know that either 1% or 10% of the launches will fail, and we have no further information on what will happen, so we assign a probability of 0.5 to both rates of failure. Because either one or the other has to be the case (by the unrealistic setup of this problem), the unconditional probability of a failure is P(F) = 0.5(0.01) + 0.5(0.10) = 0.055
Now, we see a successful launch. Intuitively, this is more likely if the rate of failure is lower, so this should favor, for the rate R, the hypothesis R = 0.01 and decrease the probability that R=0.1:
P(R = 0.01|S) = P(S|R =0.01)P(R = 0.01)/P(S) = 0.99(0.5)/0.945 = 0.524 And since it has to be one or the other, P(R = 0.1|S) = 1 − 0.524 = 0.476