aishwaryav12 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!)
\[100\,\,{\text{balls}}\,\,\,\left\{ \begin{gathered}
\,b\,\,{\text{blue}}\,\,\,\,\,,\,\,\,\,\,\,1 \leqslant b \leqslant 49\,\,\,\left( * \right) \hfill \\
\,100 - b\,\,{\text{yellow}} \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,? = b\]
\[\left( 1 \right)\,\,\,\,\frac{{b + 4}}{{100}} > \frac{1}{2}\,\,\,\,\,\,\, \Rightarrow \,\,\,\, \ldots \,\,\,\,\, \Rightarrow \,\,\,b > 46\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,b = 47 \hfill \\
\,{\text{Take}}\,\,b = 48 \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{INSUFF}}.\]
$$\left( 2 \right)\,\,\,\left( {{\rm{blue}}\,\,{\rm{:}}\,\,{\rm{yellow}}} \right)\,\,\,{\rm{removed}}\,\,{\rm{ratio}}\,\,\,\, = \,\,\,\,\left( {{\rm{blue}}\,\,{\rm{:}}\,\,{\rm{yellow}}} \right)\,\,\,{\rm{original}}\,\,{\rm{ratio}}\,\,\,\,$$
\[\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {b,y} \right)\,\,{\text{original}} = \left( {20,80} \right)\,\,\,\,\,\left[ {\frac{{20}}{{100}} = 20\% } \right]\,\,\,\,\,\,\,AND\,\,\,\,\,\left( {b,y} \right)\,\,{\text{removed}} = \left( {10,40} \right)\,\,\,\,\,\,\left[ {\frac{{10}}{{50}} = 20\% } \right]\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,b = 20\,\,{\text{viable}}\,\, \hfill \\
\,{\text{Take}}\,\,\left( {b,y} \right)\,\,{\text{original}} = \left( {40,60} \right)\,\,\,\,\,\left[ {\frac{{40}}{{100}} = 40\% } \right]\,\,\,\,\,\,\,AND\,\,\,\,\,\left( {b,y} \right)\,\,{\text{removed}} = \left( {20,30} \right)\,\,\,\,\,\,\left[ {\frac{{20}}{{50}} = 40\% } \right]\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,b = 40\,\,{\text{viable}}\,\,\,\, \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\text{INSUFF}}.\]
$$\left( {1 + 2} \right)\,\,\,\,\,\left\{ \matrix{
\,b = 47\,\,\,\, \Rightarrow \,\,\,y = 53\,\,\,\, \Rightarrow \,\,\,{{47} \over {100}} \ne {x \over {50}}\,\,\,\,,\,\,\,\,x\,\,{\mathop{\rm int}} \hfill \cr
\,b = 48\,\,\,\, \Rightarrow \,\,\,y = 52\,\,\,\, \Rightarrow \,\,\,{{48} \over {100}} = {x \over {50}}\,\,\,\left( { = {{2x} \over {100}}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,x = 24\,\,\left( {{\mathop{\rm int}} } \right)\,\,\,,\,\,\,{\rm{viable}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{only}}\,\,\,{\rm{survivor!}} \hfill \cr
\,b = 49\,\,\,\, \Rightarrow \,\,\,y = 51\,\,\,\, \Rightarrow \,\,\,{{49} \over {100}} \ne {x \over {50}}\,\,\,\left( { = {{2x} \over {100}}} \right)\,\,\,,\,\,\,\,x\,\,{\mathop{\rm int}} \hfill \cr} \right.$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
