duahsolo 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%
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.
_________________
Fabio Skilnik ::
GMATH method creator (Math for the GMAT)
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