Quote:
Lily has only red and blue balls in a jar. If she gets three balls randomly, is the probability of getting all red balls greater than the probability of getting at least one blue ball?
(1) The number of red balls is more than three times the number of blue balls.
(2) Less than (1)/(4) of the balls in the jar are blue.
Source:
Expert GlobalNice problem!
\(\operatorname{P} \left( {{\text{all}}\,\,3\,\,{\text{red}}} \right)\,\,\,\,\mathop > \limits^? \,\,\,\operatorname{P} \left( {{\text{not}}\,\,{\text{all}}\,\,3\,\,{\text{red}}} \right)\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\operatorname{P} \left( {{\text{all}}\,\,3\,\,{\text{red}}} \right) = \boxed{\frac{{C\left( {r,3} \right)}}{{C\left( {r + b,3} \right)}}\,\,\,\,\mathop > \limits^? \,\,\,\frac{1}{2}}\)
\(\left( {1 + 2} \right)\,\,\)
\(Take\,\,\left( {r,b} \right) = \left( {4,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\frac{{C\left( {4,3} \right)}}{{C\left( {5,3} \right)}} = \,\frac{1}{{10}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle\)
\(Take\,\,\left( {r,b} \right) = \left( {5,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\frac{{C\left( {5,3} \right)}}{{C\left( {6,3} \right)}} = \,\frac{{10}}{{20}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\,\,\,{\text{but}}\,\,\,{\text{increasing}}\,\,...\)
\(Take\,\,\left( {r,b} \right) = \left( {6,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\frac{{C\left( {6,3} \right)}}{{C\left( {7,3} \right)}} = \,\frac{4}{7}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\,\)
(Keeping the number of blue balls fixed and increasing the number of red ones, we know the probability of getting 3 reds must increase...)
The answer is (E), because we were able to BIFURCATE (1+2)!
The above follows the notations and rationale taught in the GMATH method.
Regards,
fskilnik.
_________________
Fabio Skilnik ::
GMATH method creator (Math for the GMAT)
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