Latest TCS Aptitude Question SOLUTION: two cards are drawn from a from a pack of well shuffled card find the probability that one is club and other is king

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

Asked In TCS semanti dey (11 years ago)

Unsolved Read Solution (14)

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Read Solution (14) :

ankit singhal  (11 years ago)

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

Is this solution Helpfull? Yes (19) | No (15)  

abhilasha  (11 years ago)

{(13c1*4c1)-1c1)/52c2}
=51/1326
ans=1/26

Is this solution Helpfull? Yes (18) | No (2)  

shivesh  (11 years ago)

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

Is this solution Helpfull? Yes (15) | No (9)  

Arka Laha  (11 years ago)

ans-->((13/52)*(3/39))+((4/52)*(12/48))=1/26

Is this solution Helpfull? Yes (6) | No (4)  

praba  (11 years ago)

p(king)=4c1
P(club)=12c1(as one king have been chosen in above 4 king cards)
total=4c1*12c1/(52c2)
=1/34

Is this solution Helpfull? Yes (3) | No (3)  

shalini saini  (11 years ago)

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

Is this solution Helpfull? Yes (1) | No (3)  

Premkumar  (11 years ago)

total pro=52c2=1326
one is club=13c1=13
another one is king=3c1=3
p=13*3/1326
p=39/1326=1/34

Is this solution Helpfull? Yes (1) | No   

semanti dey  (11 years ago)

which one is r8?bcs there are two cases..king may be a club may nt be a club..

Is this solution Helpfull? Yes (1) | No   

anand kumar  (8 years ago)

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

Is this solution Helpfull? Yes (1) | No   

yuva prasanth  (11 years ago)

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

Is this solution Helpfull? Yes | No (2)  

sudhanhsu srivastava  (11 years ago)

(13/52)*(3/51)

Is this solution Helpfull? Yes | No (1)  

Ish Mishra  (10 years ago)

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

Is this solution Helpfull? Yes | No (1)  

anil  (8 years ago)

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

Is this solution Helpfull? Yes | No   

Rashi Chauhan  (7 years ago)

(4C1*13C1)/52C2=1/17

Is this solution Helpfull? Yes | No   

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