## Thursday, 29 June 2017

### To Maximize the Probability of Red Marble

You have 50 red balls and 50 blue balls, you have to place these balls in two containers in such a way that probability of a person picking up a red ball from any container is maximum.

1. For container 1,
You should have to place one red marble into 1st Container. By this when we select this Container 1 P(Red|C1) will be 1.

2.  For container 2,
Rest all marbles should have to place into 2nd container. By this when we select this Container 2 P(Red|C2) will be 49/99.

Here, P(C1) = P(C2) = 1/2 as there is only 2 containers.

So answer will be P(Red) = P(C1)*P(Red|C1) + P(C2)*(Red|C2)
P(Red) = (1/2)*(1) + (1/2)*(49/99)
P(Red) = 0.747474.

Proof of above given solution using the Java code:
public class MaximizeRedBallProbability {
/**
* This is the code which maximize the probability.
* @param red
* @param blue
* @return probability
*/
private static double getProbability(int red, int blue) {
double max=0;
int maxRed = 0;
int maxBlue = 0;

for (int i = 1; i < red; i++) {
for (int j = 0; j < blue; j++) {
double prb = (0.5*i)/(i+j) + (0.5*(50-i))/(100-i-j);

if (prb>max) {
max = prb;
maxRed=i;
maxBlue=j;
}
}
}
System.out.println("Container 1, Red balls: "+maxRed +" Blue balls : "+maxBlue);
System.out.println("Container 2, Red balls: "+(red-maxRed) +" Blue balls : "+(blue-maxBlue));
return max;
}

/** Driver Method. */
public static void main(String[] args) {
int red = 50;
int blue = 50;
double p = getProbability(red,blue);
System.out.println("Maximum probability of RED ball "+p);
}
}
Output: 7474747474747475