Aptitude - Probability - Discussion
Discussion Forum : Probability - General Questions (Q.No. 9)
9.
From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?
Answer: Option
Explanation:
Let S be the sample space.
| Then, n(S) = 52C2 = | (52 x 51) | = 1326. |
| (2 x 1) |
Let E = event of getting 2 kings out of 4.
 n(E) = 4C2 = |
(4 x 3) | = 6. |
| (2 x 1) |
| P(E) = |
n(E) | = | 6 | = | 1 | . |
| n(S) | 1326 | 221 |
Discussion:
32 comments Page 2 of 4.
Gaurav Garg said:
1 decade ago
@Santhu.
We are not selecting cards from only 4 kings, we are selecting them from all 52 cards And next time from 51 cards.
We are not selecting cards from only 4 kings, we are selecting them from all 52 cards And next time from 51 cards.
Jhansi sri said:
1 decade ago
How can we know the cards are only kings in 52. In 4 kings we take 2 only so we take 4p2. Please help me?
Pavan Mitra said:
1 decade ago
@Jhansi shri.
We are not taking 2 kings out of four but 2 kings out of the whole deck i.e. 4c1=4c52 then we take p(one king out of 51) since we have taken 1 king out of 52 i.e. 3c51 so then P(A and B) = P(A) x P(B|A)=4/52*3/51.
We are not taking 2 kings out of four but 2 kings out of the whole deck i.e. 4c1=4c52 then we take p(one king out of 51) since we have taken 1 king out of 52 i.e. 3c51 so then P(A and B) = P(A) x P(B|A)=4/52*3/51.
Dhruvil said:
1 decade ago
Probability = (4*3/52*51) = (3/661).
As there are 4 kings in a single pack of cards and 2 cards are selected from 52 cards so the answer is (3/661).
As there are 4 kings in a single pack of cards and 2 cards are selected from 52 cards so the answer is (3/661).
Bex said:
10 years ago
Wouldn't the simple non combinational way be more simple that is probability of drawing a king is 4/52.
The probability of drawing another king is 3/51 = 4/51*3/52 = 12/2651 = 1/221 as well?
The probability of drawing another king is 3/51 = 4/51*3/52 = 12/2651 = 1/221 as well?
Naveenaa said:
9 years ago
Could anyone kindly clarify at what type of situations we are about to use this nCn?
Mahi said:
9 years ago
Prince.
Formula for nCr = n!/(n-r)!/r!
Formula for nCr = n!/(n-r)!/r!
Tiger said:
9 years ago
Solve it for me please.
An ordinary deck of 52 cards contains 4 aces 4 queens 4 kings and 4 jacks. Two cards are chosen randomly the probability that at least one ace is there.
Please give me the solution.
An ordinary deck of 52 cards contains 4 aces 4 queens 4 kings and 4 jacks. Two cards are chosen randomly the probability that at least one ace is there.
Please give me the solution.
Vinayak said:
8 years ago
There are 12 kings in 52 cards.
Then how to use 4 out 2.
There is 12 out of 2.
Then how to use 4 out 2.
There is 12 out of 2.
Ernie said:
8 years ago
@Tiger.
P(at least one ace)=1-P(not ace) ----> Complement Rule,
P(not ace) = ((52-4)C2) --->52-4 is the number of cards that are not aces,
--------------- ---> fraction bar
(52C2) ---> get 2 cards out of 52,
P(not ace) = (48C2)/(52C2) = 188/221,
P(at least one ace) = 1-(188/221) = 33/221,
@Vinayak
There are only 4 kings out of a standard deck of 52 cards; one for hearts, one for diamonds,
one for clubs and one for spades. Then, it is correct if we say, we get 2 kings (of any suit) out of the
4 kings (one per suit), (4C2).
P(at least one ace)=1-P(not ace) ----> Complement Rule,
P(not ace) = ((52-4)C2) --->52-4 is the number of cards that are not aces,
--------------- ---> fraction bar
(52C2) ---> get 2 cards out of 52,
P(not ace) = (48C2)/(52C2) = 188/221,
P(at least one ace) = 1-(188/221) = 33/221,
@Vinayak
There are only 4 kings out of a standard deck of 52 cards; one for hearts, one for diamonds,
one for clubs and one for spades. Then, it is correct if we say, we get 2 kings (of any suit) out of the
4 kings (one per suit), (4C2).
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