MBA
Exam
Tthere 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 1) 60 2) 50 3) 30 4) 80
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- 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 - 9 years agoHelpfull: Yes(0) No(0)
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