binomial distribution trials updating, is there a Bayesian solution? (Daughter's homework, first year uni, context is not clear, just basic PDf/Cdf calculations.)  An initial drug trial has known success p=0.4. ... there are some questions about this... then a final question:  A new similar treatment is trialled on 200 people and has success p=0.475.  In light of this information do you think the new programme could have a success ratio greater than 45%?  Why?
I interpret this as a sharper question: What is the probability the new treatment actually has a success p>0.45 (90 out of 200 say)?
In my dopey head I read this as P("p>0.45"|D)  where "D"="the data" = "95 out of 200 successes".  So then is this not simply,
1-pbinom(90,200,0.475) = 0.7377192  ?
(using the R syntax so pbinom is the cumulative Cdf.)
My worry was this did not use the information that a likely Bayesian prior estimate for p was p=0.4, and why else would the question hint at the fact this is the "new programme" seemingly implying we do know about an older trial with p=0.4?
How (or could?) you get a better idea of whether p>0.45, or not, using Bayesian inference?
 A: You don't give the level of your daughter's course or the context of this question within the course (e.g., recent topics studied). As stated, this does not seem to be a question in Bayesian inference. 
The size of the initial trial that gave $p = .4$ is not given. It is possible
that this is intended to be sort of a 'standard' against which results of the (presumably smaller) trial with $n = 200$ is to be tested.
If so, maybe you'd be testing $H_0: p = .45$ against $H_1: p > .45,$  using 95 Successes out of 200 subjects as data. (Admittedly, the choice of '0.45' seems arbitrary--here and in any other path of analysis.)
Another possibility would be to find a CI for $p$ based on 95  Successes in 200 and see whether it includes 0.45. 
In any case, a sample of size $n = 200$ does not seem to be quite large enough o to make fine
distinctions among $p = .4, .45, .475.$ The estimated standard error based on $\hat p = 0.475$ is $\sqrt{\hat p(1-\hat p)/200} \approx 0.16.$
The following output from Minitab statistical software does the one-sided hypothesis test I suggest above and gives the bound for the corresponding one-sided 95% confidence interval. (It uses exact binomial probabilities, not
normal approximations--hence the word 'Exact'.)
Test and CI for One Proportion 

Test of p = 0.45 vs p > 0.45

                                              Exact
Sample   X    N  Sample p  95% Lower Bound  P-Value
1       95  200  0.475000         0.414993    0.261

A relevant computation in R statistical software is as follows:
1 - pbinom(94, 200, .45)
## 0.2608107

