We have a total of 100 balls and probability of selecting blue ball is more than the probability of selecting a yellow ball.
$$P\left(Y\right)>P\left(B\right)$$
Statement 1= If four yellow balls have been repainted blue, it will have been more likely to select a blue ball rather than a yellow one.
From this statement, there are three possible variants
Yellow balls (Y)=53 and Blue balls (B)=47
$$P\left(Y\right)=\frac{53}{100}\ and\ P\left(B\right)=\frac{47}{100}$$
if we repaint four yellow balls to blue:
$$P\left(Y\right)=\frac{49}{100}\ and\ P\left(B\right)=\frac{51}{100}$$
next variants: Y=52, B=48 and Y=51, B=49.
Statement 1 is INSUFFICIENT.
Statement 2= After the representative removes fifty balls, the probability of selecting a blue ball is the same as it as it was initially
This statement says that we remove 50 balls and received the same probability this means that we removed equal quantity of of both colours hence it can include any variant that includes even numbers
Y=60, B=40, Y=72 ; B=28, ; Y=64, B=36; e.t.c. ; hence it is not sufficient
combining two statements together
from statement 1 we have 3 variants ; two odd numbers and one with even numbers and from statement 2 we know that we should have even numbers
Y=52, B=48 ;
$$P\left(Y\right)=\frac{52}{100}\ and\ P\left(B\right)=\frac{48}{100}$$
and if we repaint 4 balls the probabilities will be;
$$P\left(Y\right)=\frac{48}{100}\ and\ P\left(B\right)=\frac{52}{100}$$
and if we repaint 25balls of each colour we will have
$$P\left(Y\right)=\frac{26}{50}\ and\ P\left(B\right)=\frac{24}{50}$$

The two statement together are SUFFICIENT
$$Answer=option\ C$$

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.