L02.7 Total Probability Theorem

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6 жыл бұрын

MIT RES.6-012 Introduction to Probability, Spring 2018
View the complete course: ocw.mit.edu/RES-6-012S18
Instructor: John Tsitsiklis
License: Creative Commons BY-NC-SA
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Пікірлер: 15

@cycla 3 ай бұрын

Thank you professor Tsitsiklis!

@SecretEscapist Жыл бұрын

Thanks a lot for this amazing lecture ☺

@richardgamor9959 4 жыл бұрын

Very understandable

@louerleseigneur4532 3 жыл бұрын

Thanks MIT

@SD-uf4ng 2 жыл бұрын

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

@therasmataz2168 Жыл бұрын

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)

@berkcancosar6313 4 жыл бұрын

Can someone explain why he says weighted average at 3:52? We are not dividing by 3 so it should not be average?

@hugol354 4 жыл бұрын

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)

@berkcancosar6313 4 жыл бұрын

Hugo L yeah i got it thx a lot

@SD-uf4ng 2 жыл бұрын

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

@hugol354 2 жыл бұрын

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

@tarunpahuja3443 2 жыл бұрын

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