Quote:
Originally Posted by useral
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?

The cases that
would be different from
is when either, but not both, of these events happen:
, or
. Hope this helps.