Lucky2783 wrote:

At the beginning of a show, a lottery representative put one hundred balls in a bag. Any ball is equally likely to be selected.

Each ball is either blue or yellow. If someone picks one ball at random, it probably will be a yellow one.

How many blue balls has the representative put in the bag?

(1) If four yellow balls had been repainted blue, it would have been more likely to select a blue ball rather than a yellow one.

(2) After the representative removes fifty balls, the probability of selecting a blue ball is the same as it was initially.

Beautiful problem! (Be careful to avoid coming to wrong conclusions!)

\(100\,\,{\rm{balls}}\,\,\,\left\{ \matrix{

\,b\,\,{\rm{blue}}\,\,\,\,\,,\,\,\,\,\,\,1 \le b \le 49\,\,\,\left( * \right) \hfill \cr

\,100 - b\,\,{\rm{yellow}} \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,? = b\)

\(\left( 1 \right)\,\,\,\,{{b + 4} \over {100}} > {1 \over 2}\,\,\,\,\,\,\, \Rightarrow \,\,\,\, \ldots \,\,\,\,\, \Rightarrow \,\,\,b > 46\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\left\{ \matrix{

\,{\rm{Take}}\,\,b = 47 \hfill \cr

\,{\rm{Take}}\,\,b = 48 \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{INSUFF}}.\)

\(\left( 2 \right)\,\,\,\left( {{\rm{blue}}\,\,{\rm{:}}\,\,{\rm{yellow}}} \right)\,\,\,{\rm{removed}}\,\,{\rm{ratio}}\,\,\,\, = \,\,\,\,\left( {{\rm{blue}}\,\,{\rm{:}}\,\,{\rm{yellow}}} \right)\,\,\,{\rm{original}}\,\,{\rm{ratio}}\,\,\,\,\)

\(\left\{ \matrix{

\,{\rm{Take}}\,\,\left( {b,y} \right)\,\,{\rm{original}} = \left( {20,80} \right)\,\,\,\,\,\left[ {{{20} \over {100}} = 20\% } \right]\,\,\,\,\,\,\,AND\,\,\,\,\,\left( {b,y} \right)\,\,{\rm{removed}} = \left( {10,40} \right)\,\,\,\,\,\,\left[ {{{10} \over {50}} = 20\% } \right]\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,b = 20\,\,{\rm{viable}}\,\, \hfill \cr

\,{\rm{Take}}\,\,\left( {b,y} \right)\,\,{\rm{original}} = \left( {40,60} \right)\,\,\,\,\,\left[ {{{40} \over {100}} = 40\% } \right]\,\,\,\,\,\,\,AND\,\,\,\,\,\left( {b,y} \right)\,\,{\rm{removed}} = \left( {20,30} \right)\,\,\,\,\,\,\left[ {{{20} \over {50}} = 40\% } \right]\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,b = 40\,\,{\rm{viable}}\,\,\,\, \hfill \cr} \right.\)

\(\left( {1 + 2} \right)\,\,\,\,\,\left\{ \matrix{

\,b = 47\,\,\,\, \Rightarrow \,\,\,y = 53\,\,\,\, \Rightarrow \,\,\,{{47} \over {100}} \ne {x \over {50}}\,\,\,\,,\,\,\,\,x\,\,{\mathop{\rm int}} \hfill \cr

\,b = 48\,\,\,\, \Rightarrow \,\,\,y = 52\,\,\,\, \Rightarrow \,\,\,{{48} \over {100}} = {x \over {50}}\,\,\,\left( { = {{2x} \over {100}}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,x = 24\,\,\left( {{\mathop{\rm int}} } \right)\,\,\,,\,\,\,{\rm{viable}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{only}}\,\,\,{\rm{survivor!}} \hfill \cr

\,b = 49\,\,\,\, \Rightarrow \,\,\,y = 51\,\,\,\, \Rightarrow \,\,\,{{49} \over {100}} \ne {x \over {50}}\,\,\,\left( { = {{2x} \over {100}}} \right)\,\,\,,\,\,\,\,x\,\,{\mathop{\rm int}} \hfill \cr} \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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