DSE 2011 QUESTION 36

[eco'17]

http://economicsentrance.weebly.com/uploads/1/1/0/5/1105777/2011-option-a.pdf

while i know that the answer to this question is 0.7474
and this happens because we put 1 green ball in first box and the rest in the second box, but I'm not able to understand that how we got to this answer?
I mean I used the total probability formula and differentiated it with respect to the no. of red and green balls in box 1(say x& y respectively), but I'm not able to get this answer.
I don't where I'm going wrong, is it because of calculation mistake or my approach is wrong?

[Abhitesh]

B1 ---> g, r
B2 ---> g'=50-g, r'=50-r

P= P(G|B1).P(B1) + P(G|B2).P(B2) = 0.5 { g/(g+r) + 50-g/(100-g-r) }

Case 1 - g=r
P= 0.5

Case 2 - g<r
=> g/(g+r) < 1/2.
Let r=g+k => P(G|B1) = g/(2g+k) ; it'll be maximum when k attains least value which is k=1.
So for B1 we have r=g+1. --- (1)
In B2 g'>r'
P(G|B2) = g'/(g'+r') ; it'll be maximum when r'=0 and g'>0 --- (2)
(1) and (2) gives r'=0 => r= 50 => g = 49 => g'=1.

PS - If you differentiate P w.r.t g and r you'll get g=r as critical point. But you also need to check second derivatives. In that case, the second derivative test will be inconclusive as Pgg.Prr - Pgr^2 = 0. Furthermore, the variable in this question are discrete so be discreet when using calculus in such questions.

[eco'17]

thanks a ton!