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.

Would this be considered a 500 or 600 level question?

Brian,

Thanks for the response, that is very reassuring.

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.

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)



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...

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.

Source: Veritas Prep Free Question Bank

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.