ram draws a card randomly among card 1-23 and keep it back.then sam draws a card among those.wht is the prob than sam has drawn a card greatr than ram

• Sum of probability of drowing a card greater than drawn by ram is (22/23)+(21/23)+.....(1/23)=253/23
  but probability for choosing 1 card out of 23 is 1/23
  therefore total probability is (253/23)*(1/23)=(11/23)-ans
• 9 years agoHelpfull: Yes(19) No(1)
• 253/529
  22+21+20+........1=253
  ans is 253/23*23
• 9 years agoHelpfull: Yes(8) No(2)
• ram can draw 1 or 2 or 3 or 4......or 23 => p(ram draws 1 card)=1/23
  now for shayam can have choices to draw card greater than ram
  if ram draws card 1 then shayam has 22 choices so p(E)=1/23*22/23
  " " " " 2 " " " 21 " so p(E)=1/23*21/23
  so on........
  finally when ram draws card 22 then shayam has 1 choice so p(E)=1/23*1/23
  so total probability will be sum of all individual events
  i.e p(E)=1/23(22/23+21/23+......+1/23)
  =>1/23^2(22+21+20+.....1)
  =>1/23^2*((22*(22+1)/2)
  =>11/23 ans
• 7 years agoHelpfull: Yes(0) No(0)