from a pack of 52 playing cards,three cards are drawn at random.find the probability of drawing a king,a queen and a jack.?
explanations please

• 3P1*(4/52)*(4/51)*(4/50)=16/5525
• 10 years agoHelpfull: Yes(31) No(12)
• Ans is 4*4*4/52c3=64/22100
• 10 years agoHelpfull: Yes(27) No(6)
• There would be six possiblities for a king queen and jack...
  1st possiblity will be (KQJ)
  2nd possiblity will be (JQK)
  3rd possiblity will be (QJK)
  4th possiblity will be (QKJ)
  5th possiblity will be (JKQ)
  6th possiblity will be (KJQ)
  for drawing queen the problality will be 4/52
  for drawing jack the probablity will be 4/51
  for drawing king the probablity will be 4/50
  there are total 6 possible combination so
  6*(4/52*4/51*4/50)=16/5525
• 10 years agoHelpfull: Yes(20) No(0)
• no. of king in card=4
  no. of queen in card=4
  no. of jack in card=4
  so no. of event= 4c1*4c1*4c1=64
  no of sample space=52c3=52*17*50
  so probability=64/(52* 17*50)
• 10 years agoHelpfull: Yes(12) No(2)
• ans=6*(4/52)*(3/51)*(2/50)
  =6/5525.
  
  by arun sharma prob i of 31ques.
• 10 years agoHelpfull: Yes(8) No(4)
• ans is 16/5525
  (4c1)*(4c1)*(4c1)/52c3
  = 64/22100
  =16/5525
  
• 10 years agoHelpfull: Yes(7) No(0)
• (4/52)*(4/51)*(4/50)
• 10 years agoHelpfull: Yes(3) No(1)
• ans=4420
  
  4c1*4C1*4c1/52c3;
  
• 10 years agoHelpfull: Yes(2) No(3)
• 
  no. of king in card=4
  no. of queen in card=4
  no. of jack in card=4
  so no. of event= 4c1*4c1*4c1=64
  no of sample space=52c3=52*17*50
  so probability=64/(52* 17*50)
• 10 years agoHelpfull: Yes(2) No(1)
• from a pack of 52 playing cards, three cards are drawn at random.
  
  without any replacement:
  
  P(K,Q,J)= 12/52*11/51*10/50 (12 face cards in the deck, 3 in each suit)
  
• 10 years agoHelpfull: Yes(1) No(1)
• ans is
  (12/52)*(8/51)*(4/50)
• 10 years agoHelpfull: Yes(1) No(0)
• in a pack of 52 cards there are 4 kings,4 queens,4 aces,4 jacks...so to select one king ,one queen,one jack from 52 cards..by applying combinations we would get-4c1*4c1*4c1/52c3.
• 10 years agoHelpfull: Yes(1) No(0)
• king queen and jack can be selected as
  KQJ+KJQ+QKJ+QJK+JKQ+JQK
  4/52*4/51*4/50 ........ 6 TIMES
  SO (4/52*4/51*4/50)*6
• 10 years agoHelpfull: Yes(1) No(0)
• 4c1*4c1*4c1/52c3
• 10 years agoHelpfull: Yes(1) No(0)
• we have 4 of king, queen. and jack so, (4C1 * 4C1 * 4C1)/(52C3)=16/5525
  
  
• 10 years agoHelpfull: Yes(1) No(0)
• n(s)=52
  n(E)=4c3
  n(e)/n(s)
• 10 years agoHelpfull: Yes(0) No(1)
• probability=(4/52)*(4/51)*(4/50)=8/16575
• 10 years agoHelpfull: Yes(0) No(3)
• 3*4/52*4/51*4/50=8/5525 the pure answer
• 10 years agoHelpfull: Yes(0) No(0)
• 3*4/52*4/51*4/50
• 10 years agoHelpfull: Yes(0) No(0)
• (4c1*4c1*4c1)/(52c3)
• 10 years agoHelpfull: Yes(0) No(0)
• ans is:-6/5525
  the same question is given in arunsharma LOD 1 Q.NO-30.page:-572
• 9 years agoHelpfull: Yes(0) No(0)
• 6/5525
  arun sharma LOD1 Q.NO-30(Ph no:-572)
• 9 years agoHelpfull: Yes(0) No(2)
• 12c1 *8c1 *4c1=12*8*4;
  n(s)=52c3= 52*51*50=5525
  p(e)= 12*8*4/52*51*50
  =====>16/5525
  
• 8 years agoHelpfull: Yes(0) No(0)
• The first two answers used combinations. You can also use the laws of probability. First we calculate the probability that the first card is an ace, the second a king and the third a jack: 4/52 * 4/51 * 4/50 = 64/132600. (The second and third factors are conditional probabilities). It is easy to see that the same probability applies to any order of drawing and we may add these probabilities. As there are 6 possible orders, the answer is 6 * 64/132600 = 64/22100 = 16/5525.
• 8 years agoHelpfull: Yes(0) No(0)
• Deck of cards = 52 cards. There are 4 Aces (A), 4 Kings(K) & 4 Jacks(J) in the deck.
  
  Three cards drawn can be AJK, AKJ, KAJ, KJA, JAK, JKA (That's 6 combinations)
  
  Probability = 6 x 4/52 x 4/51 x 4/50
  
  = 384/132600 which simplifies to 16/5525
• 7 years agoHelpfull: Yes(0) No(0)