Ram draws a card randomly among cards numbered 1-23 and keep it back.Then sam draws a card among those then what is the probality that Sam has drawn a card greater than Ram.


Options
1) 84/142
2) 84/143
3) 84/144
4) 84/145

• 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.
  
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  .please somebody explain ......above question is somehow different from this one??????????
  solution for this question :
  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
• 9 years agoHelpfull: Yes(11) No(3)
• 84/143 is the answer
  probab of ram drawing a card =1/23....
  then sam drawing a card have 21 cases
  like if ram chooses card with no 1
  then sam can have card no's 2,3,4,5......23 this is one case ie 1/23 *22/23.
  same id ram chooses card with no 2 the sam shold have chosen 3,4,5,6....23
  so next case will be 1/23*21/23..... then the final equation will arise as
  1/23[22/23+21/23+20/23+.......2/23] so the answer will be 84/143....
  CORRECT ME IF I M WRONG>>>>
• 9 years agoHelpfull: Yes(8) No(8)
• Ans is 84/143
• 9 years agoHelpfull: Yes(2) No(2)
• a) 13^4 * 48 *47 is the correct answer.
  
  ways of choosing 1 card from hearts is 13C1
  ways of choosing 1 card from hearts is 13C1
  ways of choosing 1 card from diamonds is 13C1
  ways of choosing 1 card from spades is 13C1
  Rest 2 cards can be from any of the suits. So, 48C1 and 47C1
  Total no. of ways that we have cards from all suits with the chosen 6 cards is
  
  13C1*13C1*13C1*13C1*48C1*47C1 = 13*13*13*13*48*47= 13^4*48*47
• 9 years agoHelpfull: Yes(1) No(1)
• how did u get ans 84/143 arpit dubey.. plz solve it.
• 9 years agoHelpfull: Yes(1) No(2)
• http://www.campusgate.co.in/2015/11/elitmus-previous-questions-detailed-solutions-2.html
  check the ans from above link
• 6 years agoHelpfull: Yes(1) No(0)
• i think the series will be like this...check out..1/23*22/23 + 1/22*21/22 +1/21*20/21 . . . . . . +1/2*1/2 .........i guess you know why i have written this..cause ram is keeping it back thus decrementing the cards to choose frm in subsequent turns.but how the series will be solved :O ??
• 9 years agoHelpfull: Yes(0) No(5)