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MATH REVIEW PROBABILITY

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MATH REVIEW PROBABILITY  [#permalink]

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New post 22 Feb 2019, 10:36
“Can someone explain me the reason behind this?? I can’t figure it out,it’s taken by the official guide math review btw” P(A)=0.23 P(B)=0.40 P(C)=0.85 Suppose events A and B are mutually exclusive and B and C are independent.
Note that P(A or C) and P(A and C) cannot be determined using the given information. But it can be determined that A and C ARE NOT mutually exclusive since P(A)+P(C)=1.08 which is greater than 1, and therefore cannot equal P(A o C). From this it follows that P(A and C)>/ P(C) =0.85 since C is a subset of A U C. Thus, one can conclude that 0.85<\ P(A or C) <\ 1 and 0.08<\ P(A and C) <\ 0.23

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Re: MATH REVIEW PROBABILITY  [#permalink]

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New post 22 Feb 2019, 13:44
Hi, saramacchia

I think you are talking about this one. please have a look at this discussion.
https://gmatclub.com/forum/gmat-og-13-probability-questions-173028.html
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Re: MATH REVIEW PROBABILITY   [#permalink] 22 Feb 2019, 13:44
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MATH REVIEW PROBABILITY

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