Finance Majors/Non-Finance Majors - Overlapping Set Question

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In a given finance lecture, 30% of the students are finance majors, and 40% of the students are female. The gender distribution for finance majors and non-finance majors is the same. If one student is called on at random, what is the probability that the student is neither female nor a finance major?

A. 70%
B. 60%
C. 58%
D. 42%
E. 30%

I have become pretty adept at using the double-set matrix. But the wording "the gender distribution for finance majors and non-finance majors is the same" threw me off.

OA: D

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by GMATGuruNY » Tue Jan 29, 2013 7:23 pm
alexander.vien wrote:In a given finance lecture, 30% of the students are finance majors, and 40% of the students are female. The gender distribution for finance majors and non-finance majors is the same. If one student is called on at random, what is the probability that the student is neither female nor a finance major?

A. 70%
B. 60%
C. 58%
D. 42%
E. 30%
The gender distribution for ALL the students = 40% female, 60% male.
For the gender distribution for finance majors and non-finance majors to be the same:
Finance majors = 40% female, 60% male.
Non-finance majors = 40% female, 60% male.
(If the gender distribution is ANY OTHER RATIO -- if both types of majors are 30% female, 70% male, for example -- then the gender distribution for all the students will NOT be 40% female, 60% male.)

We can use the following formula for overlapping groups:

Total = Group 1 + Group 2 - Both + Neither.

The big idea with overlapping groups is to SUBTRACT THE OVERLAP.
When we count everyone in Group 1 (finance students) and everyone in Group 2 (females), those who are in BOTH groups (female finance students) get counted twice.
So that we don't double-count those who are in both groups, we SUBTRACT THE OVERLAP from the total.

In the problem at hand:
Let the total = 100.
Group 1 = finance majors = 30.
Group 2 = females = 40.
Since 40% of the finance majors are female, BOTH female and a finance major = .4(30) = 12.
Let N = the number of students who are NEITHER female NOR a finance major.
Plugging these values into the equation above, we get:

100 = 30 + 40 - 12 + N
N = 42.

Thus, P(neither female nor a finance major) = 42/100 = 42%.

The correct answer is D.
Last edited by GMATGuruNY on Tue Jan 29, 2013 8:15 pm, edited 1 time in total.
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by bpolley00 » Tue Jan 29, 2013 7:56 pm
Would this be considered a 500 or 600 level question?

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by brianlange77 » Tue Jan 29, 2013 8:52 pm
bpolley00 wrote:Would this be considered a 500 or 600 level question?
Gut instinct... high 500/no more than low 600.

The math here is pretty straight-forward.

Hope this helps.

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by bpolley00 » Wed Jan 30, 2013 8:05 am
Brian,

Thanks for the response, that is very reassuring.

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by cking6178 » Fri Feb 01, 2013 1:26 pm
For overlapping sets, isn't it easier/quicker to derive the answer by multiplying the probabilities? I arrived at 42% by multiplying the probability of Non Finance (70%) by the probability of Male (60%), so .6*.7 = .42. These are the questions that should help you gain extra time on the harder problems.

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alexander.vien wrote:In a given finance lecture, 30% of the students are finance majors, and 40% of the students are female. The gender distribution for finance majors and non-finance majors is the same. If one student is called on at random, what is the probability that the student is neither female nor a finance major?

A. 70%
B. 60%
C. 58%
D. 42%
E. 30%
As Alexander notes, this question can be solved using the Double Matrix method. For those who are unfamiliar with this method, here's how it looks:

Note: This technique can be used for most questions featuring a population in which each member has two criteria associated with it.
Here, the criteria are:
- Major (Finance or Non-Finance)
- Gender (Female or Male)

Since we're dealing with percents all the way through to the answer choices, let's make things easy on ourselves and say that there are 100 students in the lecture.
So, here's the setup.
Image

30% of the students are finance majors
So, 30 students are finance majors and 70 are not.
Image

40% of the students are female
We get:
Image

The gender distribution for finance majors and non-finance majors is the same.
In other words, there's a 40/60 female/male split among the finance majors and among the non-finance majors.
We get:
Image

What is the probability that the student is neither female nor a finance major?
In other words, what is the probability that the student is a male non-finance major?
Once we simplify the boxes . . .
Image
. . . we see that 42 of the 100 students meet this criteria.
So, the probability = 42/100 = [spoiler]42% [/spoiler]= D

For a more detailed explanation information of the Double Matrix method and some additional practice questions, check out these 3 BTG articles:

- https://www.beatthegmat.com/mba/2011/05/ ... question-1
- https://www.beatthegmat.com/mba/2011/05/ ... question-2
- https://www.beatthegmat.com/mba/2011/05/ ... question-3

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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I have been a big fan of double matrix method, it simplifies any overlapping set problem (ensure columns need to be mutually exclusive - many thanks to Brent for patiently explaining this)

I set my table this way (since we are given that the gender distribution for F and Non F is same - easier to visualize a ratio)

Image

Now, you can plug in the value in ~F / Male box.
You can eliminate 70 and 60 right away (if 70, then M / F will be negative, so eliminate and if 60, then M / F = 0)
A quick glance will tell you 58 isn't right either, since 58/12 <> 3/2 (gender ratio)

42 works [spoiler](42 / 28 = 3/2 and to double check, sub values in first column, which would be 18 and 12 conforming to 3/2 ratio) [/spoiler]

So, double matrix to set up and then quick plug in of numbers, if another option to solve.

My 2 cents...

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by fskilnik@GMATH » Wed Oct 17, 2018 1:48 pm
alexander.vien wrote:In a given finance lecture, 30% of the students are finance majors, and 40% of the students are female. The gender distribution for finance majors and non-finance majors is the same. If one student is called on at random, what is the probability that the student is neither female nor a finance major?

A. 70%
B. 60%
C. 58%
D. 42%
E. 30%

I have become pretty adept at using the double-set matrix. But the wording "the gender distribution for finance majors and non-finance majors is the same" threw me off.

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Image

VERY beautiful problem, in which the Venn diagram ("overlapping sets") solves the problem nicely.
$$? = {\rm{None}}\,\,\,\,\left( {\,{\rm{in}}\,\,100\,} \right)$$
The wording in red is clearly presented below and also in the diagram!
$${{{\rm{finance}}\,{\rm{majors}}\,\,{\rm{and}}\,\,{\rm{female}}} \over {{\rm{finance}}\,{\rm{majors}}}} = {{{\rm{non - finance}}\,{\rm{majors}}\,\,{\rm{and}}\,\,{\rm{female}}} \over {{\rm{non - finance}}\,{\rm{majors}}}}\,\,\,\, \Rightarrow \,\,\,\,{x \over {30}} = {{40 - x} \over {70}}$$
$${x \over 3} = {{40 - x} \over 7}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,7x = 120 - 3x\,\,\,\,\, \Rightarrow \,\,\,\,x = 12$$
$${\rm{finance}}\,{\rm{majors}}\,\,{\rm{or}}\,\,{\rm{female }}\,{\rm{ = }}\,\,30 + 40 - x\,\,\mathop = \limits^{x\, = \,12} \,\,\,58\,\,\,\,\,[\,{\rm{simplifier}}\,]$$
$$? = {\rm{Total}} - 58\,\,\mathop = \limits^{{\rm{Total}}\, = \,100} \,\,\,42\,\,\,\,\,\,\,\left( {42\% } \right)$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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