Probability_Balls_problem

Question

There are $2$ boxes, $1$ containing $10$  red balls and the other containing $10$ green balls.

You are allowed to move the balls between the boxes so that when you choose a box at random and a ball at random from the chosen box, the probability of getting a red ball is maximized.

What is the maximized Probability?

Answer 1

The probability for selecting red box is 1/2 but randomly if we select green box, the probability will be 0.
so, total probability for this case is 0.5.

now consider the other case, 9 red balls are transferred to green ball box. now total balls in that box will be 19.

total probability = probability to select 1st box + probability to select 2nd box

= (1/2)*(1C1/1C1)+(1/2)*(9C1/19C1)

= (1/2)*(1+9/19)

= (1/2)*(28/19)

= 14/19

Comments

@ Avdhesh Singh Rana

now consider the other case, 9 red balls are transferred to green ball box. now total balls in that box will be 19.

 why are u considering only 9 balls....i think to maximize the chance we must transfer all(10) the red balls ...

and probality would be

 (1/2)*1 + (1/2)*(10C1/20C1) ->1/2 +1/4 ->3/4

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Once you transfer all 10 balls from one to another, then there will be one empty bucket.
1/2 (random bucket to choose) * (1/2(10 red balls out of 20 balls)+0(nothing is there)) = 1/4

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...no..thinl this way...

bag A and Bag B and we know that one is containing all red balls and other is containing all green balls....

now probability of selecting bag having red ball=1/2*(probability for selecting a red ball i.e 1) +selction of other bag 1/2 then transfer all from this to other then 10C1/20C1 ....

1/2+1/4