There are two box-es, one contains 12 red balls and the other containing 47 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. This maximum probability is:

• Put 1 Red ball in Box 1
  Put the remaining 11 Red balls and the 47 Green Balls in Box 2
  Probability of selecting either box = 1/2
  Probability of taking from 1st Box=1C1
  Probability of Taking from 2nd Box=11C1/58C1 (58 means 47+11 total)
  So, Probability of selecting a red ball =
  (1/2)(1C1) (for Box 1) + (1/2)(11C1/(58C1) (for Box 2) = 1/2 + 1/2 (11/58)
  
  =0.594
• 13 years agoHelpfull: Yes(25) No(2)
• probability of choosing any box randomly = (1/2)
  for maximizing the probability of getting red ball, in the second draw we have to get red ball and its probability wiil be = (12C1)/(59C1)
  as the total no of balls are 59.
  so the maximized prob. = (1/2)*[(12C1)/(59C1)]
  = 0.102
• 13 years agoHelpfull: Yes(9) No(11)
• NOW WE DRAW A BALL AT RANDOM SO BOTH BOTH BOXES ARE HAVING THE PROBABILITY 1/2
  I.E FIRST BOX IS HAVING PROBABILITY 1/2 AND SAME SECOND ONE IS 1/2
  NOW DRAW A RED BALL FROM FIRST BOX IS 1/2(10C1/10)
  NOW THE PROBABILITY OF GETTING RED BALL IS MAXIMIZED SO WE MOVE 11 RED BALLS TO SECOND BOX THEN SECOND BOX HAVING TOTAL BALLS IS 58(47+11)
  NOW DRAW A RED BALL IS 1/2(11C1/58C1)
  PROBABILITY IS 1/2(1)+1/2(11/58)=69/116
• 13 years agoHelpfull: Yes(5) No(0)
• 1st box =1 red balls
  2nd box = 11 red balls + 47 green balls = 58 balls
  maximize red ball so combination of red ball = 12c1 from 58 balls = 58c1
  So, prob. to select 1 red ball from 58 balls is 12c1/58c1
  Prob. that 1 red ball is select in 1st box = 1c1
  No of colors is 2. so prob. to select 1 red ball = 2c1
  
  Now max prob. = (1c1+(12c1/58c1))/2c1
  =>(1+(12/58))/2
  =>(70/58)/2
  =>1.20/2
  =>0.60
• 13 years agoHelpfull: Yes(1) No(6)