Latest TCS Aptitude Question SOLUTION: four cards are drawn from a pack of 52 cards one by one without replacement. what is the probability that each card is of different suit and different domination

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four cards are drawn from a pack of 52 cards one by one without replacement. what is the probability that each card is of different suit and different domination

Asked In TCS abhishek ranjan (10 years ago)

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abhishek ranjan  (10 years ago)

no condition for drawing 1st card.
suppose drawn card is 7 of spade. the 2nd card drawn must not be either spade or 7.

favorable space = (52)-(13 cards of spade)-(3 cards of denomination "7") =36
total no of space =51
P2= 36/51

same way for 3rd card drawn-
favorable = 36-(14) =22
total =50
p3=22/50

for last card-
p4=10/49

reqd probability =(52*36*22*10)/(52*51*50*49)

Is this solution Helpfull? Yes (15) | No   

JITENDRA  (10 years ago)

suppose if you draw card '7 of spade' in 1st draw,in the 2nd draw all spades will be excluded and 7 of diamonds,hearts,club will also be excluded,so you will have only 36 cards to choose from 51.
Next draws will follow the same logic.
required ans = (52 x 36 x 22 x 10) / (52 x 51 x 50 x 49)

Is this solution Helpfull? Yes (9) | No (2)  

yanna raskala  (10 years ago)

1st card is taken from 52 cards=52/52 (of one suit and one domination)
2nd card is taken from 51 cards=36/51 (of 2nd suit and other domination)
3rd card is taken from 50 cards=24/50 (of 3rd suit and other domination)
4th card is taken from 49 cards=12/49 (of 4th suit and other domination)
so answer is (52*36*24*12)/(52*51*50*49)

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ABHISHEK KUMAR MISHRA  (10 years ago)

1st card is selected from 52 cards, I don't care what will be P1=1
for 2nd card p2=3/51
for 3rd card p3=2/50
for 4th card p4=1/49
so for all 4 cards the answer will be 1*(3/51)*(2/50)*(1/49) = 6/(51*50*49)

Is this solution Helpfull? Yes (1) | No (8)  

swati singh  (10 years ago)

there are 4 suits of 13 each.. one card is drawn from each so 4*13c1/52c4

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soujanya kolukuluri  (10 years ago)

1st card is taken from 52 cards=52/52 (of one suit and one domination)
2nd card is taken from 51 cards=38/51 (of 2nd suit and other domination)
3rd card is taken from 50 cards=24/50 (of 3rd suit and other domination)
4th card is taken from 49 cards=10/49 (of 4th suit and other domination)
so answer is (52*38*24*10)/(52*51*50*49)

Is this solution Helpfull? Yes | No (3)  

yanna raskala  (10 years ago)

1st card is taken from 52 cards=52/52 (of one suit and one domination)
2nd card is taken from 51 cards=36/51 (of 2nd suit and other domination)
3rd card is taken from 50 cards=22/50 (of 3rd suit and other domination)
4th card is taken from 49 cards=10/49 (of 4th suit and other domination)
so answer is (52*36*22*10)/(52*51*50*49)

Is this solution Helpfull? Yes | No (1)  

jaffreyhudson  (10 years ago)

(13C1)^4/52C4

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