>> Solutions to a) is p, and b) is 2p/(1-p). > > That can't be right. Try 2p/(1 + p).
Sorry about that. You are absolutely right, 2p/(1+p) it is.
> A strangely worded question. I assume that "we make an observation of > a cow or a crow, with probability q and 1-q" means this: we toss a > biased coin, or whatever, to decide whether to observe a cow or a > crow. If we need to observe a crow then we pick one crow at random > from the population of crows, and see whether it is white or black. > Why bother, one might ask, since we are told to "assume all crows are > black".
I don't find that strange. That just means that we want to verify our hypothesis, and when we go out to the jungle, we observe cows 100q percent of the time, and 100(1-q) percent of the time. Lets say we will do 30 experiments, then this model is better, than maybe having to stick around forever to wait for a crow, if 1-q is very small.
> But anyway, observing the colour of a pre-determined non-cow object > (such as a crow) can't possibly affect the probability that all cows > are white. What *would* have an effect is if we pick a non-white > object at random and find that it is not a cow (for example, it is a > crow). The key thing is that we *could* have picked a non-white cow if > one existed. Since the number of observable non-white objects is > extremely large (though presumably finite), the effect is tiny.
I am not sure I follow. But could I rephrase it as follows. If we had no information saying "all crows are black", then the two probabilities in a) and b) should coincide?