Cfrgtkky

Z is actually a probability function. I am finding where the probability is maximized. But I could find no way how to maximize this function.

Original question is as follows:

Mr A wants to join a Gamer's club. There are two identical boxes filled with Red and Green balls, and he has to pick up a green ball in order to join the club. You are required to allocate balls in such a way that the probability of A joining The Gamer's club is maximized and boxes should be non-empty. There are 50 red and 50 green balls in total. What is the maximum probability?

** I have taken the number of Green balls in Box 1 as x and red balls in Box 1 as y. And hence I have created the probability function given in the question title.**

The maximum probability occurs where there is only 1 green ball and 0 red ball in Box 1 and the value of probability is 0.747. I cannot figure a way out of how to reach this number.

All I want is a rule book method in how to proceed in such type of questions.

Introducing new variable $p=2(x+y)$ leads to:

$$f(y,p)=frac{y}{p}+frac{50-y}{200-p}$$

New variables for convenience:

$$y=50r \ p=50 s$$

$$g(r,s)=frac{r}{s}+frac{1-r}{4-s}$$

$$0 leq r leq 1, qquad 0 < s < 4$$

Now we use the usual derivative method:

$$g_r=frac{1}{s}-frac{1}{4-s}=0$$

$$g_s=-frac{r}{s^2}+frac{1-r}{(4-s)^2}=0$$

From these equations we have:

$$s_0=2 \ r_0 = frac{1}{2}$$

However, the tests show this is not a maximum. Neither it's a minimum, more likely a saddle point, though I haven't checked properly.

Since we haven't been able to find a maximum point in this region, that means that the maximum is achieved somewhere on the boundary.

And unfortunately, this simply means that we have $s to 0$ or $s to 4$ where the function becomes unbounded.