In a chess tournament of 4 games among anand and karsproviz, probability of anand wins is 2/5. Probability of individual game draw is zero. What is the probability of the series draw??(i.e.2-2)

[Question model from CAT by Arun sharma_LOD-III. Better you read all those LOD-III.. My friends got a few more 4m that only]

• probability of anand win=2/5
  probability that anand loses=probability of karsprovizwin=1-2/5=3/5
  now series will be a draw if anand and karsproviz win 2 matches each i.e aakk.
  this pattern can be arranged in 4!/2!2! = 6 ways(aakk,akak,kkaa,kaka,kaak,kaak)
  probability of individual way= (2/5*2/5*3/5*3/5) = 36/625
  required probability = 6*36/625=216/625
• 10 years agoHelpfull: Yes(25) No(1)
• IF THE SERIES IS TO BE A DRAW THEN ANAND SHOULD WIN 2 GAMES . THEREFORE
  
  4C2 (2/5)^2 (3/5)^2 = 6*4*9/625=216/625
• 10 years agoHelpfull: Yes(8) No(2)
• probability of anand=2/5
  probability of karsp0viz=1-2/5=3/5
  probability of draw =1/4*(2/5*2/5*3/5*3/5=9/625
  p=9/625
  
  explanation:1/4 for 6 case out of 24 series draw(aakk,akak,akka,kaak,kaka,kkaa)
• 10 years agoHelpfull: Yes(7) No(8)
• Cat level questions r aske din tcs ????
• 10 years agoHelpfull: Yes(6) No(0)
• x=wining chance of anand
  then P(x)=2/5
  the the chance of anand loss the game=chance of karsproviz won the game=3/5
  (because there is no chance to individual game draw)
  x~B(4,2/5)
  P(X=2)=4C2*(2/5)^2*(3/5)^2
  =216/625
  =0.3456
• 10 years agoHelpfull: Yes(3) No(0)