there are 23 chits numbered from 1 to 23. sonu pick up a chit and puts it back. monu picks another chit and puts it back. what is the probability that monu's chit number is greater than that of sonu's.

• the no of cheat is 1 2 3 4 5................23,so the ans is 1/23*22/23+1/23*21/23+.......................1/23*1/23
  1/23*1/23(22+21+20+19+....................1)
  (1/23*1/23)(22*23)/2=11/23 ans
• 9 years agoHelpfull: Yes(18) No(1)
• Easier Way : There is one possibility that both cards picked will be same.
  Removing that we are left with 22 ways of choosing a card out of which only exactly half ways are of choosing a greater(smaller) card ..i.e. 11
  That's why probability = 11/23.
  Correct me if i am wrong.
• 9 years agoHelpfull: Yes(8) No(0)
• The no of ways sonu can pick up the chits=23
  The no of ways monu can pick up the chits=23
  therefore total no of ways chits can b picked up=23*23(maximum no of cases)
  
  now if sonu picks 1 monu can pick 22 larger chits
  if he picks 2 monu can pick 21 larger chits
  and so on.....
  therefore no of favourable cases = 22+21+20......+1
  this is an ap with d=-1 a=23 n=23
  sum = 23*11
  therfore required probab=(23*11)/(23*23)=11/23
• 9 years agoHelpfull: Yes(6) No(1)
• 1/23 *( 22/22 + 21/22 + 20/22 + .... + 1/22) = 0.5
• 9 years agoHelpfull: Yes(3) No(3)
• 1/2-1/23=22/46=11/23
• 9 years agoHelpfull: Yes(0) No(2)
• The answer would be 1/2
  bcoz sonu picks a card and put it back now we left 22 cards ,total no of occurences is (23*22)
  now no of occurrence for which monu picks greater than sonu is simple....think
  (1,2),(1,3)..........(1,23)->23
  +
  (2,3),(2,4).........(2,23)->22
  .....this way no of occur=1+2+3......23=11*23
  p(n)= (11*23) /(22*23)=11/22=0.5
• 9 years agoHelpfull: Yes(0) No(1)