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# probability and distribution help

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Joined: 31 Dec 1969
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Kudos [?]: 202 [0], given: 102262

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10 Jan 2012, 19:53
explain how to do p(b|c')
i dont get it :s
given that :
P(A) = 0.70
P(B) = 0.11
P(C) = 0.30
P(AnB) = 0.05
P(AnC) = 0.21
P(BnC) = 0.06
P(AuBuC) = 0.80
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 3709
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Kudos [?]: 5857 [1] , given: 66

Re: probability and distribution help [#permalink]

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11 Jan 2012, 12:37
1
KUDOS
Expert's post
Hi, there. I'm happy to help with this.

The first thing I'll say: I have never seen a conditional probability question on the GMAT. I believe this question is a more advanced probability topic than you would see on the GMAT. It is a typical question of, for example, the AP Statistics exam.

Here's how you solve it.

First of all, we know P(C) = 0.30, so the complementary region is P(C') = 1 - P(C) = 0.70. If 30% of all members are in C, then 70% are not in C.

Next, this is a sophisticated move. If you think about the region B, it has two parts: the part the overlaps with C and the part that doesn't overlap with C. Obviously, the total region B is the sum of those two sub-regions. Thus

P(B) = P(BnC) + P(BnC') --> 0.11 = 0.06 + P(BnC') ---> P(BnC') = 0.05. If 11% of all members are in B, and 6% of all members are in both B & C, then it must be that 5% of all members are in B and not in C.

Now, the conditional probability P(B|C') (read, "the probability of B, given not C") is the ratio of all the region that comprises "B and not C" to all the region of "not C." In other words, within this situation, the condition "not C" is the "all", and we want to know what part of that "not C" universe does B occupy? The formula is:

P(B|C') = P(BnC')/P(C') = (0.05)/(0.70) = 1/14 = 0.07142857

A Google search for the term "conditional probability" will turn up curricula galore if you want more information or example problems. I will stress, though: all this advanced probability stuff is way beyond what the GMAT will demand of you.

Please let me know if you have any questions about what I've said.

Mike
_________________

Mike McGarry
Magoosh Test Prep

Re: probability and distribution help   [#permalink] 11 Jan 2012, 12:37
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