Unit 3 10a Answer
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@kousik04 Yeag, it's clear now. I found some other explanation, but thanks to u too.
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@kousik04 Sorry i mean, P(A AND B) is not ALWAYS the same as P(A given B has occurred).
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@artivenom The basic flaw in your argument is P(A AND B) is not the same as P(A given B has occurred).
In fact, P(A and B) = P(A given B) * P(B). So, if A and B are independent events (true in both your case as well as Thrun's question), then P(A and B) = P(A)*P(B). So equating the two, we get P(A)*P(B) = P(A given B) * P(B).
Cancelling out P(B), we get P(A given B) = P(A) if A and B are independent events.
Hope this clarifies.
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what the heck......
I really be fooled by this question...
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explained before why not? My brain went into ~TILT~
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@artivenom It's basically a nonsensical question. P(R|S) = P(R) because R is independent of S. (And take this with a grain of salt, because I found this video after working through 10-15 lines of calculation and giving up.)
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how is it possible?
1. box has 70/100 black balls
2. box has 1/100 black ball.
What is a probability of taken the two balck balls (The words "given that it's sunny" says that), if I take a black ball from first box AND black ball from the second box after that? 0.7* 0.01 because I take the second ball from the box AFTER the first one(or at the same time). And we don't need the result only black ball from second one, but from all of the boxes. Or not?
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This answer link is not showing on the website, ai-class, FYI.
puppetrhapsody 4 months ago 12
This doesn't appear on the website
df747jet 4 months ago 2