#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? 
#2




Re: Hw 2, problem 3
Quote:
__________________
When one teaches, two learn. 
#3




Re: Hw 2, problem 3
Thanks, that clarifies it.

Thread Tools  
Display Modes  

