two cards are drawn from a from a pack of well shuffled card find the probability that one is club and other is king

• total prob: 52C2= 52*51/2=26*51
  prob of getting club is= 13C1= 13
  prob of getting king is= 4C1=4
  ans= 13*4/26*51= 2/51
• 12 years agoHelpfull: Yes(19) No(15)
• {(13c1*4c1)-1c1)/52c2}
  =51/1326
  ans=1/26
  
• 12 years agoHelpfull: Yes(18) No(2)
• probablity of club=13/52
  probability of king=4/51
  so answer =13/52*4/51
  but if 1st card is king of club then
  13/52*3/51
  so answe=91/2652
• 12 years agoHelpfull: Yes(15) No(9)
• ans-->((13/52)*(3/39))+((4/52)*(12/48))=1/26
• 12 years agoHelpfull: Yes(6) No(4)
• p(king)=4c1
  P(club)=12c1(as one king have been chosen in above 4 king cards)
  total=4c1*12c1/(52c2)
  =1/34
  
• 12 years agoHelpfull: Yes(3) No(3)
• total prob=52c2
  prob of selecting club is 13c1
  prob of selecting king is 4c1
  therefore prob of getting one club and other as king is 13c1*4c1/52c2
  2/51
  
• 12 years agoHelpfull: Yes(1) No(3)
• total pro=52c2=1326
  one is club=13c1=13
  another one is king=3c1=3
  p=13*3/1326
  p=39/1326=1/34
• 12 years agoHelpfull: Yes(1) No(0)
• which one is r8?bcs there are two cases..king may be a club may nt be a club..
  
• 12 years agoHelpfull: Yes(1) No(0)
• let 2 cases firstly in club when we r taking 1 card then king is not taken
  13c1x4c1/52c2
  other, when we r taking 1 card from spade then that card is king. so, we have only 3 kngs now and we have to take 1 out of 3king 13c1x3c1/52c2
  add both 2/51 +3/162 =1/26
  
• 9 years agoHelpfull: Yes(1) No(0)
• 4/13..ie..13/52 for club and 3 other kings are there in remaining cards...so that will be 3/52...so totally 4/13
• 12 years agoHelpfull: Yes(0) No(2)
• (13/52)*(3/51)
  
• 12 years agoHelpfull: Yes(0) No(1)
• Let X be the event that cards are in a club which is not king and other is the king of club.
  Let Y be the event that one is any club card and other is a non-club king.
  Hence, required probability:
  =P(A)+P(B)
  =C112×C11C252+C113×C13C252
  =(2×(12×1)52×51)+(2(13×3)52×51)=(24+7852×51)
  = 1/26
• 11 years agoHelpfull: Yes(0) No(1)
• Let X be the event that cards are in a club which is not king and other is the king of club.
  
  Let Y be the event that one is any club card and other is a non-club king.
  
  Hence, required probability:
  
  =P(A)+P(B)
  
  =12C1×1C152C2+13C1×3C152C2
  
  =(2×(12×1)52×51)+(2(13×3)52×51)
  
  =(24+7852×51)
  
  =1/26
• 9 years agoHelpfull: Yes(0) No(0)
• (4C1*13C1)/52C2=1/17
• 8 years agoHelpfull: Yes(0) No(0)