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
a)60 b)50 c)80 d)30

• Well Keep a single red ball in 1 box and move 38 red balls to the other
  
  1st box = 1R
  2nd box = 38R + 26G
  
  p(red) = 1/2 * 1 + 1/2 * 38/64 ≈ 0.8
• 11 years agoHelpfull: Yes(28) No(2)
• the probability of getting a red ball should be maximized so we will keep 1 red ball in the first box and transfer remaining 38 red balls to other box which has green. The reason for doing this only this combination will make the probability of getting red ball is maximum (i.e) 1.
  In the question its mentioned that first we have to choose a box at random.
  So 2 boxes are there, probability of selecting a bag is 1/2
  Probability of getting red ball from first box is 1
  Probability of getting green ball from second box is 38/64
  So (1/2)*1 for first box and (1/2)*(38/64) for box 2
  answer is (1/2)*1 + ((1/2)*(38/64) which is approximately 0.8
• 11 years agoHelpfull: Yes(12) No(2)
• chockkalingam s pls write correctly 38/64 is probability of getting red not green.
• 11 years agoHelpfull: Yes(2) No(2)