L02.7 Total Probability Theorem 

Not correct. we are adding each probability of B with respect to an Ai that contributes the total probability of B. The Ai's are collectively exhaustive so you can imagine some Aj that does not have some B in it the contribution to B's total probability would be zero although its weight would be P(Aj)

it's not the average as we usually think of, but it's the average of the event P(B/Ai) happening(he says weighted because each P(Ai) is already weighted because they sum to 1)

Hugo L yeah i got it thx a lot

​@@hugol354 weight should be p(B/Ai) and probability should be p(Ai) cause sum of p(Ai)=1 isnt it?

@@SD-uf4ng no, it is the weights on the conditional prob that B will happen knowing that A happens

@@hugol354 I think probability of a scenario is already a fraction so need not to divide.