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Re: Set Theory! [#permalink]
P(A u B u C)= P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + 2 P(A n B n C) - atleast one
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Re: Set Theory! [#permalink]
sudhir18n, can you please clarify?

If I solve your equation using the equations above, I would get

P(A u B u C)= P(A n B n C)

Which is definatley not true. What am I missing here?
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Re: Set Theory! [#permalink]
bellcurve wrote:
sudhir18n, can you please clarify?

If I solve your equation using the equations above, I would get

P(A u B u C)= P(A n B n C)

Which is definatley not true. What am I missing here?


use this
https://grockit.com/blog/gmat/2011/01/28 ... et-theory/
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Re: Set Theory! [#permalink]
The link says the same thing. I have the formulas, its just that I am not sure why formula for P(AuBuC) is different than "at least one." Aren't those refering to the same things? It seems like, "at least one" is obtained by adding P(AnBnC) to P(AuBuC). What can be included in the "at least one" that can not be included in P(AuBuC)?
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Re: Set Theory! [#permalink]
Can anybody shed some light?
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Re: Set Theory! [#permalink]
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bellcurve wrote:
Anybody?


I think that 'at least one' should be the same as union. But they seem to have something else in mind (I don't have the faintest clue what that is). In terms of the diagram, the area that corresponds to 'at least one set' according to them, doesn't represent anything.
You may want to ask them for clarification on their blog post.



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