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# Gmat Prep Probability question

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Intern
Joined: 21 Sep 2006
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Gmat Prep Probability question [#permalink]  02 Oct 2006, 10:24
Hello all -

Just completed a practice session using the GmatPrep software and since ETS doesn't provide solutions, I am hoping someone can help solve/explain the steps to getting the solution. Please help! Here goes:

In a survey of 248 people, 156 are married, 70 are self employed, and 25 percent of those who are married are self employed. If a person is to be randomly selected from those surveyed, what is the probability that the person selected will be self employed but not married?

a) 1/8 b) 4/31 c) 117/248 d) 1/4 e) 31/117

Last edited by \$uckafr33 on 02 Oct 2006, 10:38, edited 1 time in total.
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SVP
Joined: 05 Jul 2006
Posts: 1516
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Kudos [?]: 124 [0], given: 39

First i would ( if u excuse me ) to draw ur attention to that it is much better for u and all of us on the forum to post ur questions one at a time ( one question in each post)........hope u dont mind
SVP
Joined: 05 Jul 2006
Posts: 1516
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Kudos [?]: 124 [0], given: 39

FOR QUESTION ONE MY ANSWER IS A ...kindly find attached...it will help u
Attachments

Intern
Joined: 11 Nov 2005
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Yup.. got A as my answer as well.
Senior Manager
Joined: 13 Sep 2006
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Location: New York
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One more for A.

248 surveyed

25% of 156 = 39 (total married and seld employed).

70 = total self employed - 39 (total married and self employed) = 31

31/248 is 1/8 (don't make the silly mistake i orginally did thinking 31 had no factors and that something was wrong with final calc!).
Director
Joined: 28 Dec 2005
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total = 248
se = 70
^se=178
m=156
^m=92
se & m = 39
=>se & ^m = 31

P(se & ^m) = (se & ^m)/ total = 31/248 = 1/8
Intern
Joined: 21 Sep 2006
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I too got all the way up to 31 and thought to myself that it was a prime number and thus unfactorable.

Great job everyone!
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