A teacher gives three bags to abhishek, amitabh and aiswarya.Each bag contains six balls. abhishek has 2 red balls, amitabh has 5 red balls.When a ball is drawn by them the probability of not being same colour is 5/6.the number of red balls in aiswarya bag is ?
a)2 b)3 c)4 d)5

• ans- a)2
  suppose aish has x red balls.
  
  and given that 1 ball is drawn by them, i.e, 1 ball is drawn from each bag
  then prob of not being same color is 5/6
  
  now first let me find out prob of being same color: ((2c1*5c1*xc1) + (4c1*1c1*(6-x)c1)/6*6*6
  then prob of not being same color is: 1- (2*5*x + 4*1*(6-x))/6*6*6
  so 1- (2*5*x + 4*1*(6-x))/6*6*6 = 5/6
  solve this equation, x=2
  
  hence Aish contains 2 red balls in her bag.
  
  
• 10 years agoHelpfull: Yes(68) No(6)
• key +> ball can be any ao three bag and probability of getting ball which is not reed;
  
  probability of selecting a ball from any one of three bag = 1/3;
  probability of selecting a ball which is not red = 4/6+1/6+(6-x)/6;
  where x is no of red ball in ash bag;
  1/3[4/6+1/6+(6-x)/6] = 5/6;
  => x= 4;
• 10 years agoHelpfull: Yes(7) No(6)
• this type of problems go through options wise
  so take first bag(abhi)-2 red balls,assume remains are 4blue
  take second bag(amit)-5 red balls,assume ramains one is green
  take third bag(ais)-x red balls ,assume remains y blue yellow
  in the problem each bag have six balls and also we have not being same color probability i.e 5/6
  write a not being same color probability
  bag1 (or) bag2 (or) bag3 (drawn one ball only see the problem)=5/6
  2/6 (or) 1/6 (or) y/6 = 5/6
  red (or) green (or) yellow
  (2+1+y)/6 =5/6
  if y=2 then only condition is satiefeid(condition is not satiesfied remains values 3,4,5 )
  here 2-balls is yellow not red balls,so remains(6(total balls)-2(yellow))=4red balls
  ans:C.
  pls reply this procedure correct or not
• 10 years agoHelpfull: Yes(5) No(0)
• anyone solve dis???
• 10 years agoHelpfull: Yes(1) No(2)
• correct answer is option b) 3
• 10 years agoHelpfull: Yes(1) No(3)
• anyone solve dis???
• 10 years agoHelpfull: Yes(0) No(3)
• anyone solve dis???
• 10 years agoHelpfull: Yes(0) No(2)
• anyone solve dis???
• 10 years agoHelpfull: Yes(0) No(2)
• if you have trouble finding it u can also manipulate with the options.
• 10 years agoHelpfull: Yes(0) No(0)
• I think option
• 9 years agoHelpfull: Yes(0) No(0)