For the last question
If we assume the test result is deterministic, then
P(D2=1|D1=1) = 1
P(D2=0|D1=0) = 1
Doing first experiment twice does not add additional information. Therefore, P(H=1|D1=1,D2=1) == P(H=1|D1=1). You can derive the equation by doing some arithmetic.
I don’t understand equation 2.6.3 . On the right side, why wouldn’t P(A) on the top cancel out with P(A) on the bottom, and since the other term on the bottom right which is the sum of all b in B for P(B|A) equals 1, wouldn’t that mean it would then just simplify to P(A|B) = P(B|A) which is obviously incorrect?
There seems to be something wrong with the typesetting.
Yes, the equation is slightly wrong. The updated equation should sum over all possible ‘a’ values in the sample space(a and its complement so that it gets normalized accurately), Reference
An error in exercise 4? It says we draw n samples and then uses m in the definition of zm.
Would have been great and more clear to actually see what numbers you multiplied in the example, for someone never doing stats before, it can be hard to comprehend all the formulas without any explanations.