I'm not sure if it's directly applicable to your situation, but here is a brief discussion on choosing
an informative prior in a beta-binomial context.
I'm not sure how much experience you have with Bayesian
inference, so my apologies if this is way too simple.
Let $\theta$ be the relapse probability, which you
believe to be around 11%. It seems natural to use a
beta distribution, which has support $(0,1)$ and
is conjugate with binomial data.
One beta distribution that is "more or less" concentrated around $0.11$ is $\theta \sim \mathsf{Beta}(\alpha=3, \beta=24).$ It has mean $3/27 = 0.111$ and it puts
95% of its probability in $(0.024, 0.251).$
[Computations in R.]
qbeta(c(.025, .975), 3, 24)
[1] 0.02445808 0.25130292
Then if you observe 7 relapses among 50 patients in the class being studied, your posterior distribution is
$$p(\theta|x) \propto \mathsf{Beta}(3+7 = 10,\; 24+43 = 67),$$ which has mean 0.149, median 0.127 and gives the 95% posterior probability interval $(.065, 0.213).$
[For reference: An Agresti-Coull frequentist 95% CI for the same data is $(0.067, 0.266).$ A Bayesian interval estimate based on the Jeffreys noninformative prior is $\mathsf{Beta(.5,.5)}$ is $(0.065, 0.255).]$
qbeta(.5, 10, 67)
[1] 0.1266618
qbeta(c(.025, .975), 10, 67)
[1] 0.06494157 0.21293324
If you want your prior to be more influential, then try
priors with shape parameters in the same ratio, but both larger--perhaps $\mathsf{Beta}(6, 48),$
which puts about 95% of its probability in the narrower prior interval $(0.042, 0.207).$ And so on.
qbeta(c(.025, .975), 6, 48)
[1] 0.04269639 0.20658533
Of course, with a strong prior such as $\mathsf{Beta}(60, 480),$ we would hardly pay attention to data based on only 50 patients. Posterior interval $(0.089,0.140).$
qbeta(c(.025,.975), 60+7,480+43)
[1] 0.08925147 0.14034708