There are two boxes,one containing 39 red balls & the other containing 26 green balls.you are allowed to move the balls b/w the boxes so that when you choose a box random & a ball at random from the chosen box,the probability of getting a red ball is maximized.this maximum probability is
Choices:
A) 0.9  B) 0.8  C) 0.5  D) 0.7  E) 0.2
Ans: Very interesting question.   As we are allowed to move the balls, we keep only one red ball in first box and move all the remaining balls to the second box   So fist box contains 1 redball, second box contains 38 red + 26 green = 64 balls   Probability of choosing any box is 1/ 2.   So probability of taking one red ball = 12×(1)+12(3864)≃0.8   Answer:0.8
<result>B</result>
There are two boxes,one containing 39 red balls & the other containing 26 green balls.you are allowed to move the balls b/w the boxes so that when you choose a box random & a ball at random from the chosen box,the probability of getting a red ball is maximized.this maximum probability is