There are two boxes, one containing 32 black balls and other containing 31 red balls you are allowed to move the boxes at random so that when you choose the box at random and a ball at random from the chosen box, the probability of getting the black ball is maximized. This maximum probability is
(a).25
(b).51
(c).75
(d).50

• keeping one black ball in a box and move rest to the box containing red balls. So now selecting a black ball from the two balls is as :-
  selecting black from first box =1/2*1C1=0.5
  selecting black from 2nd box=1/2*(31C1)/(62C1)=0.25
  therfore probability =0.5 +0.25=0.75
  
  
• 8 years agoHelpfull: Yes(5) No(0)
• Move 31 balls from black ball box to other box having 31 red balls.
  
  now one box has one black ball
  and
  other box is having 31 black balls and 31 red balls.
  
  then prob of getting black ball = 1/2 (1+31/62) = 1/2 *3/2 =3/4=0.75
• 8 years agoHelpfull: Yes(5) No(0)
• choosing one box=1/2
  prob of black=32c1
  tot prob=63c1
  req prob=1/2*(32c1)/(63c1)
  =.25
• 8 years agoHelpfull: Yes(3) No(1)
• 0.50 ans
  
  
• 8 years agoHelpfull: Yes(2) No(5)
• you have choose 1 box from 2 boxes so prob is 1/2
  among 63 ball u r going to choose 1 ball hence prob is 63c1
  for getting black ball prob is 32c1
  so 32c1/(63c1*0.5)=0.254
• 8 years agoHelpfull: Yes(2) No(3)
• Number of Black balls=32
  Number of Red balls=31
  Total balls=66
  Probalility to pick one black ball from total balls=32/66=>0.50
• 8 years agoHelpfull: Yes(2) No(8)
• max prob= prob of choosing 1 box (1/2) + prob of choosing black ball ( 32c1/63C1)=0.75 ans
• 8 years agoHelpfull: Yes(2) No(0)
• some one can tell which ans is correct? please
• 8 years agoHelpfull: Yes(1) No(3)
• they asked to move the boxes and not the balls..so I guess we are not supposed to interchange the balls..keeping this in mind max probability of putting the hand in box containing black balls is 0.50
  
• 7 years agoHelpfull: Yes(1) No(0)
• Probability of chosen the box is =1/2
  and now we have to calculate the selecting red ball from the box
  i,e. (1/2)*31/62=1/4
  now the probability of selecting black ball is 1-1/4
  i.e. 3/4=.75
• 7 years agoHelpfull: Yes(1) No(0)
• Move 31 balls from black ball box to other box having 31 red balls.
  now one box has one black ball and other box is having 31 black balls and 31 red balls.
  
  then prob of getting black ball = 1/2 (1+31/62) = 1/2 *3/2 =3/4 =.75
• 7 years agoHelpfull: Yes(1) No(0)
• Ans- 1/2 * 1 + 1/2 * 31/ 63 = 0.75
• 8 years agoHelpfull: Yes(0) No(2)
• Move 31 balls from red ball box to other box having 31 green balls.
  
  now one box has one red ball
  and
  other box is having 31 red balls and 31 green balls.
  
  then prob of getting red ball = 1/2 (1+31/62) = 1/2 *3/2 =3/4
• 8 years agoHelpfull: Yes(0) No(0)
• so ,total outcomes =63,favourable outcomes of black=32,probability=0.50 the answer is d
• 8 years agoHelpfull: Yes(0) No(0)
• A : B : C = 3 : 2 : 5
  (33) Means 3, Therefore 33/3 = 11 .
  3*11=33
  2*11=22
  5*11=55
  -------------
  + 110 (Ans)
• 7 years agoHelpfull: Yes(0) No(2)