#1




*ANSWER* Hw 2, problem 3
Hello,
I must be missing the forest for the trees, but I was thinking that the answer should be the weighted sum of the probabilities of (y = f(x)) and (y =/= f(x)), and the weight mu (corresponding to an error, i.e. y =/= f(x)) should be assigned to the probability of that error, i.e. to 1lambda. This reasoning gives the answer [d]. Is it flawed? 
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