Two cards are drawn at random from a pack of 52 cards. What is the probability that the drawn cards

P.Pandeeswari   10 years AGO

probability that the both cards are same=4P2/52P2=4*3/52*51=1/221
option 8)1/221

Like? Yes (27) | No (4)  

Udhaya kumar   10 years AGO

4C2 / 52C2 = 1 / 221........

Like? Yes (14) | No (4)  

santosh kumar patel   10 years AGO

n(S) = 52C2 = 1326
n(E) = 4c2 = 6
P(E) =n(E)/n(S) = 1/221

Like? Yes (11) | No (2)  

Sravanthi   10 years AGO

(4/52)*(3/51)=1/221

Like? Yes (8) | No (1)  

rosemary   10 years AGO

There are 4 aces in a 52-card deck. Suppose without replacement, the probability is
P(2 aces)= (4/52) * (3/51) = (1/13) * (1/17) = 1/221

Like? Yes (7) | No   

Anoop Tiwari   10 years AGO

for first card drawn probability of ace=4/52=1/13
for second card drawn probability of ace=3/51
so total probability=1/13* 3/51=1/221
hence option 8) is correct

Like? Yes (5) | No   

padmaja   10 years AGO

4c2/52c2
=1/221

Like? Yes (4) | No   

Swagatam Kumar Das   10 years AGO

Probability of getting two aces is 4c2
So ans is 4c2/52c2=1/221.

Like? Yes (4) | No   

Manasa Dullur   10 years AGO

4C2/52C2 = 1/221

Like? Yes (3) | No   

Parag Wagh   10 years AGO

4c2/52c2=1/221

Like? Yes (3) | No   

swarupa nuggu   10 years AGO

1/221

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Ramesh Sriram   10 years AGO

4c2/52c2

Like? Yes (5) | No (2)  

Aparna Chakraborty   10 years AGO

4/52 *3/51 =1/221

Like? Yes (3) | No   

Himanshu Srivastava   10 years AGO

4c2/52c2

Like? Yes (2) | No   

aman   10 years AGO

There r 4 aces
So 4c2
Total card 52
52c2
Prob 4c2/52c2 = 1/221

Like? Yes (2) | No   

vinit   10 years AGO

probabilities: 4/52*3/51=1/221

Like? Yes (2) | No   

A SIRISHA    10 years AGO

the way to draw 2 cards out of a possible 4 aces = 4P2
the way to draw 2 cards from a deck of 52 cards= 52P2
4P2/52P2=1/221

Like? Yes (2) | No   

Jayanta Dutta   10 years AGO

4 aces can be selected as 4C2 And from 52 we can select as 52C2 so prob is-4C2/52C2
=1/221

Like? Yes (3) | No (1)  

Anubhav jain   10 years AGO

4/52*3/51
1/17*1/13
=1/221

Like? Yes (2) | No   

G.Praveen Kumar   10 years AGO

there are 4 aces so while drawing 1 card pfobility is 4/52 and while drawing 2nd 3/51
taking both (4/52)*(3/51)=1/221

Like? Yes (2) | No   

Jabez David   10 years AGO

There are 4 aces in a 52-card deck.

Suppose without replacement, the probability is

P(2 aces)= (4/52) * (3/51) = (1/13) * (1/17) = 1/221

Like? Yes (2) | No   

Anshul Shrivastava   10 years AGO

(4/52)*(3/51)=(1/221)

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Surit Datta   10 years AGO

probability = 4C2/52C2 = 1/221
That's it..

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swati kunj   10 years AGO

probability of getting aces=4c2/52c2 =1/221

Like? Yes (2) | No   

SUCHETA BISWAS   10 years AGO

4c2/52c4=1/221

Like? Yes (3) | No (2)  

laxmi kumari   10 years AGO

Let S be the sample space . Then
n(S)=1326
Let E be the event of getting 2 aces out of 4. Then
n(E)=6
so,n(P)=1/221

Like? Yes (3) | No (2)  

Supriya Mondal   10 years AGO

We know that there is 4 ace card in a deck of playing card..
so probability is=4c2/52c2=1/221

Like? Yes (1) | No   

ravi prakash   10 years AGO

4c2/52c2= 1/221

Like? Yes (1) | No   

saikumar.chejerla   10 years AGO

4c2/52c2

Like? Yes (1) | No   

Karthick Raja   10 years AGO

4 aces in a deck of cards
first ace=4/52
second ace=3/52
p(two aces)=4/52*3/51
=12/2652
=1/221

Like? Yes (1) | No   

Vivek Raj   10 years AGO

4/52*3/51=1/221

Like? Yes (1) | No   

kancharla kavya   10 years AGO

Since there are only 4 ace cards in a pack of 52 cards ....Then the probability that both are aces is 4c2/52c2=1/221....

Like? Yes (1) | No   

Chirag Ranjan Lahiri   10 years AGO

Clearly, there are 52 cards, out of which there are 4 aces.

P (getting both aces) = 4c2/52c2 = 12/2562 = 1/221 answer.......

Like? Yes (2) | No (1)  

Md.Sultan Khichi   10 years AGO

4C2/52C2=1/221

Like? Yes (1) | No   

SUNITEE GARIMA   10 years AGO

P=4C2/52C2=1/221

Like? Yes (1) | No   

Aparna Bhattacharya   10 years AGO

Let S be the sample space and E be the the event of getting 2 aces out of 4. Then,
n(S)=52C2=1326
n(E)=4C2=6
P(E)=n(E)/n(S)=6/1326=1/221
Option: 8

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m.lokeswari   10 years AGO

there are four aces in the pack we select two out of four so the answer is 1/221

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MRITUNJAY   10 years AGO

proability of aces=4c2/52c2
=1/221

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dhanaseelan   10 years AGO

4C2/52C2=1/221

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Arkja tiwari   10 years AGO

4c2/52c2

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MAHARAJAN.G   10 years AGO

p(e)=n(e)/n(s)
there are four aces
n(e)=4p2
total cards is 52 drawn card is 2
n(s)=52p2
4p2/52p2=4*3/52*51=12/2652=1/221

Like? Yes (2) | No (1)  

Zaheen Shekh   10 years AGO

total card for aces is 4 so the first case is 4/52. this is first selection then sencond is 3/51. now both multiplied 4/52*3/51=1/221

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CHANDAN KUMAR   10 years AGO

4c2/52c2=1/221

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Gaurav Sharan   10 years AGO

p(E)= 4C2/52C2
i.e. (4*3/2*1)/(52*51/2*1)=1/13*17= 1/221

Like? Yes (1) | No   

niraj   10 years AGO

6/1326=1/221

Like? Yes (1) | No   

khushboo goenka   10 years AGO

4C2/52C2

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rakesh   10 years AGO

4/52*3*51=1/221

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ayushi   10 years AGO

probability of drawing 2 aces is= 4C2/52C2=1/221

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vaitesh mudaliar   10 years AGO

total no. of outcomes is 52C2.
no. of ways through which we can take out exactly 2 aces from 4 ace cards is 4C2. probability = 4C2/52C2 =1/221

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Shivan kumar jha   10 years AGO

4c2/52c2'
4c2=6
52c2=1326
6/1326=1/221
opction==8

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Riyaz   10 years AGO

probability of getting an ace is 4/52 and getting an other ace is 3/51. so probability of getting 2 aces is 4/52*3/51=1/221

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MD SHAMIM MONDAL   10 years AGO

4C2/(52C2)
there are 4 aces out of which 2 are taken so 4C2
there are 52 card a whole out of which 2 are taken

Like? Yes (1) | No   

RAMAKRISHNA   10 years AGO

p(ace 1)=4/52
p(ace 2)=3/51
p(both are ace)=(4/52)*(3/51)=1/221

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Akash Verma   10 years AGO

4c2/52c2
since first selecting 2 cards of ace from total of 4 aces and as we are selecting 2 cards of total 52 cards
so,, 4c2/52c2=1/221 is the answer

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Chandra chur anand   10 years AGO

probability= (4c2) / 52c2= 1/221

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bhavyasri nunna   10 years AGO

2 ace cards are selected from 4 ace cards so p(ace/total) is 4c2/52c2

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A M ANKIT KALLURAYA   10 years AGO

PROBABILITY=NO OF FAVOURABLE CASES/TOTAL NO OF CASES
=(4/52)(3/21)=1/221

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sachin verma   10 years AGO

selection of 2 aces from 4 aces=4c2
and selection of 2 cards from 52 cards=52c2
so P(E)=4c2/52c2=1/221

Like? Yes (1) | No   

Arijit Acharyya   10 years AGO

(4/52)*(3/51) = 1/221

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Amit Karmakar   10 years AGO

probability = (4*3)/(52*51)
= 1/221

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anjali s   10 years AGO

4 aces,selceting 2 =4C2 /52c2(2 from 52 cards)

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LOKESH KUMAR   10 years AGO

I'm assuming that you are NOT replacing the first card and then drawing the second. For the first card drawn , probability of ace is 4/52. Then for the second card, after the first ace is drawn, there are only 3 aces left out of 51 cards, so probability is 3/51. The probability of performing two independent activities such as this is actually the product of the probabilities:

Like? Yes (1) | No   

Akshay   10 years AGO

(4/52)*(3/51)=1/221

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Deeksha Sharma   10 years AGO

4C252C2=6/1326=1/221

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Santosh Kumar Behera   10 years AGO

Let S sample space then n(S)= 52C2 =(52*51)/2*1=1326

Let E is the getting of Aces 2 out of 4
n(E)=4C2=6
P(E)=n(E)/n(S)=6/1326=1/221

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JILCY J ABRAHAM   10 years AGO

Answer:option 8
If we are not replacing the first card and then drawing the second, the solution is given below.
There are 4 aces out of 52 cards in the packet.
When the first card drawn, the probability of ace is 4/52.then we take the second card ,the probability of getting an ace is 3/51.
The probability of performing two independent activities such as this is actually the product of the probabilities
I.e. (4/52)*(3/51) = (1/13) (1/17)
=1/221

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Sourabh Prasad   10 years AGO

(4C2)/(52C2)

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Naga Divya Peddi   10 years AGO

n(E):aces are 4 from out of 52 cards..so only 2 are drawn
so 4C2
now n(S)=52C2 n now P(E)=n(E)/n(S)
i.e 4c2/52c2=6/1326=1/221
so option is 8

Like? Yes (1) | No   

Hemadri   10 years AGO

n(s)=52C2=1326
n(E)=4c2=6
P(E)=n(s)/n(E)=6/1326=1/221

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Sandeep Kumar   10 years AGO

4/52 * 3/51 = 1/221

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BHASKAR MANDAL   10 years AGO

4c2 / 52c2 = 1/221...option 8

Like? Yes (2) | No (2)  

abhishek narayan   10 years AGO

4c2/52c2=1/221

Like? Yes (2) | No (2)  

bhagya   10 years AGO

probability of ace = 4/52
Then for the second card=3/51
(4/52)*(3/51)
=(1/13)*(1/17)
=1/221

Like? Yes (1) | No (1)  

P.ARUNA   10 years AGO

4c2/52c2
=6/26x51
1/221

Like? Yes (1) | No (1)  

Chandra Bhushan Kumar Dubey   10 years AGO

4C2/52C2=1/221

Like? Yes (2) | No (2)  

ARJUN PRASAD   10 years AGO

4c2/52c2

Like? Yes (2) | No (2)  

Deepak prajapati   10 years AGO

4c2/52c2=1/221

Like? Yes (2) | No (2)  

v pavan   10 years AGO

probability of drawing 1st ace= (4/52)
probability of drawing 2nd ace= (3/51)
total probability= (4/52)*(3/51)

Like? Yes (1) | No (1)  

Abhishek mani   10 years AGO

4c2/52c2
=1/221

Like? Yes (2) | No (2)  

MINTU KUMAR GUPTA   10 years AGO

(4C2)/(52C2)
=1/221

Like? Yes (2) | No (2)  

Biswajit Saha   10 years AGO

(4/c2)/(52c2)=1/121

Like? Yes (1) | No (1)  

prem kumar chaudhary   10 years AGO

4c2/52c2=1/221

Like? Yes (2) | No (2)  

Tark Patel   10 years AGO

( 4 P 2 ) / ( 52 P 2) = ( 4 * 3 ) / ( 52 * 51 ) = 1 / 221

Like? Yes (1) | No (1)  

Rishikesh Ranjan   10 years AGO

4c2/52c2= 1/221

Like? Yes (1) | No (1)  

Arijit Das   10 years AGO

ToTAL NO. OF CARDS= 52
NO. OF ACES= 4

PROBABILTY OF DRAWING 2 ACES= 4C2/52C2 = 1/221

Like? Yes (1) | No (1)  

lakshmi   10 years AGO

probability of ace = 4/52
Then for the second card=3/51
(4/52)*(3/51)
=(1/13)*(1/17)
=1/221

Like? Yes (1) | No (1)  

purushothama   10 years AGO

probability of ace = 4/52
Then for the second card=3/51
(4/52)*(3/51)
=(1/13)*(1/17)
=1/221

Like? Yes (1) | No (1)  

Narasapu Hari prasad   10 years AGO

option 8 is correct
no.of ways of selecting two aces are=4c2=6
no.of ways of selecting 2 cards from 52 cards=52c2=1326
so required probability=6/1326=1/221

Like? Yes (1) | No (1)  

jayalakshmi   10 years AGO

probability of ace = 4/52
Then for the second card=3/51
(4/52)*(3/51)
=(1/13)*(1/17)
=1/221

Like? Yes (1) | No (1)  

bikash tiwari   10 years AGO

4c2/52c2

Like? Yes (2) | No (2)  

APPIKONDA SANTOSH NAIDU   10 years AGO

4c2/52c2

Like? Yes (2) | No (2)  

sanjeev kumar   10 years AGO

And is 8

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CHANDA VIJAYALAKSHMI   10 years AGO

probability of ace = 4/52
Then for the second card=3/51
(4/52)*(3/51)
=(1/13)*(1/17)
=1/221

Like? Yes (1) | No (1)  

nikhil pogale   10 years AGO

probablity=4*3/52*51
= 12/2652
= 1/221

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K.ARun   10 years AGO

4/52 * 3/51

Like? Yes (1) | No (1)  

Ramrup Chatterjee   10 years AGO

(4c2)/(52c2)
=1/221

Like? Yes (2) | No (2)  

SUBHASH TULSYAN   10 years AGO

4c2/52c2=1/221

Like? Yes (1) | No (1)  

RAJEEV RANJAN   10 years AGO

4c2/52c2=1/221

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M CHIRANJIBI   10 years AGO

this is the answer

Like? Yes (1) | No (1)  

poojasaisri   10 years AGO

there are 4 ace cards in pack of 52 cards
probability of taking 2 ace cards =
4c2/52c2=(4*3/2*1)/(52*51/2*1)=1/(13*17)=1*221

Like? Yes (1) | No (1)  

Zameer Khan   10 years AGO

4/52*3/51
=1/221

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Pravallika   10 years AGO

4c2/52c2=3/663=1/221

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himaja   10 years AGO

there are 4 aces
so probability of picking 2 aces = 4/52 + 3/51 = 1/13+1/17 = 1/221

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swarna   10 years AGO

4/52 * 3/51

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Sanu Panji   10 years AGO

For the first card drawn , probability of ace is 4/52.
Then for the second card probability is 3/51.
Total probability = (4/52)*(3/51)= 1/221

So option 8)

Like? Yes (1) | No (1)  

Mani Shankar   10 years AGO

chance of first card being ace is: 4/52,
and chance of being second card ace is: (4-1)/(52-1)
so, 4/52 * 3/ 51 = 1/ 221

Like? Yes (1) | No (1)  

sravani rachakonda   10 years AGO

4c2/52c2=1/221

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Y V S Chandrakanth   10 years AGO

4c2 / 52c2=1/221

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shubham sethiya   10 years AGO

4C2/52C2= 1/221

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Anand Dwivedi   10 years AGO

= 4·3 = 12 = 1
52·51 2652 221

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RAKESH   10 years AGO

P = 4C2 / 52C2 = 1/221

Like? Yes (1) | No (3)