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Difficult Math Problem #77 - Probability

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800guy Master | Next Rank: 500 Posts Default Avatar
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Difficult Math Problem #77 - Probability

Post Mon Dec 18, 2006 8:21 am
Find the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains:

A) exactly 1 woman
B) at least 1 woman
C) at most 1 woman

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Stacey Koprince GMAT Instructor
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Post Sat Jan 06, 2007 3:46 pm
Hey, guys

Just wanted to add to the conceptual "exactly / at least / at most" discussion.

We have five possibilities for the number of women chosen to be in a group of 4 people: 0, 1, 2, 3, 4.

The first question asks for exactly 1, so we can just calculate the probability for that option. We want 1 woman AND not 1 woman AND not 1 woman AND not 1 woman. Note that "AND" means multiply in probabilities. (See 800guy's post for the actual calculations.)

The second question asks for at least one woman. That means we want either 1 woman OR 2 women OR 3 women OR 4 women. (Note that "OR" means add in probabilities.) It's really time-consuming to calculate 4 possibilities and add them together, so we look for a shortcut. All of the probabilities must add to 1 (to cover 100% of the scenarios), and we're asked to find four out of five. So, instead, let's just find the fifth possibility (zero women) and subtract that from 1. Much faster and - even more importantly - less work to do means fewer chances to make mistakes.

The third question asks us for at most one woman, which means either 0 women OR 1 woman. Here, we actually do have to calculate the two possibilities and add them together, because the alternative - calculate the other 3 possibilities and subract them from 1 - is not a shortcut. Note that this set up rarely appears on the GMAT, primarily because there is no real shortcut here. For most of the harder math questions, there is some sort of a shortcut and part of the eventual score differentiation is whether you know it or can figure it out quickly.

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Last edited by Stacey Koprince on Mon Jan 08, 2007 4:21 pm; edited 1 time in total

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800guy Master | Next Rank: 500 Posts Default Avatar
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Post Mon Jan 08, 2007 3:51 pm
awesome explanation stacey

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800guy Master | Next Rank: 500 Posts Default Avatar
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Post Sun Dec 31, 2006 12:26 pm
OA coming when a few people reply

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maxim730 Senior | Next Rank: 100 Posts Default Avatar
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Post Tue Jan 02, 2007 11:17 am
Eagerly awaiting answer. I always wanted to know how to compute "at least""at most" "exactly"!

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thankont Senior | Next Rank: 100 Posts
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Post Wed Jan 03, 2007 7:35 am
a) exactly 1 woman
1st way (personally I like it better)

x=(7C1*11C3)/18C4

2nd way
x=7/18*11/17*10/16*9/15

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800guy Master | Next Rank: 500 Posts Default Avatar
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Post Wed Jan 03, 2007 11:55 am
OA:

A) 7C1* 11C3/ 18C4
B) 1 - (11C4/18C4)
C) (11C4/18C4) + (7C1*11C3/18C4)

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