Latest TCS Aptitude Question SOLUTION: 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 dr

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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]

Asked In TCS haran (10 years ago)

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pranav  (10 years ago)

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

Is this solution Helpfull? Yes (25) | No (1)  

SUNNY KUMAR SHAW  (10 years ago)

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

Is this solution Helpfull? Yes (8) | No (2)  

mohan lal garg  (10 years ago)

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)

Is this solution Helpfull? Yes (7) | No (8)  

ankita jain  (10 years ago)

Cat level questions r aske din tcs ????

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rajesh  (10 years ago)

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

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