Latest Elitmus Aptitude Question SOLUTION: 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

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(#M40039548) ELITMUS Numerical Ability question probality Keep an EYE

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

Asked In Elitmus amit kumar (9 years ago)

Unsolved Read Solution (25)

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KANHAIYA  (9 years ago)

3P1*(4/52)*(4/51)*(4/50)=16/5525

Is this solution Helpfull? Yes (31) | No (12)  

Narasapu Hari prasad  (9 years ago)

Ans is 4*4*4/52c3=64/22100

Is this solution Helpfull? Yes (27) | No (6)  

ALOK KUMAR  (9 years ago)

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

Is this solution Helpfull? Yes (20) | No   

vivek kumar  (9 years ago)

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)

Is this solution Helpfull? Yes (12) | No (2)  

vishvesh  (9 years ago)

ans=6*(4/52)*(3/51)*(2/50)
=6/5525.

by arun sharma prob i of 31ques.

Is this solution Helpfull? Yes (8) | No (4)  

snehal daware  (9 years ago)

ans is 16/5525
(4c1)*(4c1)*(4c1)/52c3
= 64/22100
=16/5525

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himanshu uniyal  (9 years ago)

(4/52)*(4/51)*(4/50)

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

surbhi dubey  (9 years ago)

ans=4420

4c1*4C1*4c1/52c3;

Is this solution Helpfull? Yes (2) | No (3)  

Asad Siddiqui  (9 years ago)

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)

Is this solution Helpfull? Yes (2) | No (1)  

sneh  (9 years ago)

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)

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Rachit Agarwal  (9 years ago)

ans is
(12/52)*(8/51)*(4/50)

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prem  (9 years ago)

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.

Is this solution Helpfull? Yes (1) | No   

Sahil Choudhary  (9 years ago)

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

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AMIT KUMAR  (9 years ago)

4c1*4c1*4c1/52c3

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ramanjulu  (9 years ago)

we have 4 of king, queen. and jack so, (4C1 * 4C1 * 4C1)/(52C3)=16/5525

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Deepak Dubey  (9 years ago)

n(s)=52
n(E)=4c3
n(e)/n(s)

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shanu jain  (9 years ago)

probability=(4/52)*(4/51)*(4/50)=8/16575

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Sourav sarkar  (9 years ago)

3*4/52*4/51*4/50=8/5525 the pure answer

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Rahul  (9 years ago)

3*4/52*4/51*4/50

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manjeet kumar  (9 years ago)

(4c1*4c1*4c1)/(52c3)

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Farha Masroor  (8 years ago)

ans is:-6/5525
the same question is given in arunsharma LOD 1 Q.NO-30.page:-572

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Farha Masroor  (8 years ago)

6/5525
arun sharma LOD1 Q.NO-30(Ph no:-572)

Is this solution Helpfull? Yes | No (2)  

rahul kumar  (7 years ago)

12c1 *8c1 *4c1=12*8*4;
n(s)=52c3= 52*51*50=5525
p(e)= 12*8*4/52*51*50
=====>16/5525

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amit maurya  (7 years ago)

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.

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chandni jaiswal  (7 years ago)

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

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