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 3 of 4.
Lola said:
1 decade ago
The n(k) = 4, so the p(getting a king)=4c1, or 4/52 : p(getting 2 king) = 4c2. NB: combination mean selection.
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?
Naveenaa said:
9 years ago
Could anyone kindly clarify at what type of situations we are about to use this nCn?
Alok said:
3 years ago
As drawing of cards are dependent events, I think 4c1*3c1/52c2 will be correct.
(1)
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.
Ajith kumar said:
1 decade ago
@ Shyam
Generally there are only four kings in a pack of 52 cards.
Generally there are only four kings in a pack of 52 cards.
S.Mounica said:
1 decade ago
Its formula, nC2 = n*n-1/1*2
So we get it.
So we get it.
Prince said:
1 decade ago
(52 x 51)
---------..? how is this equation possible..?
(2 x 1)
---------..? how is this equation possible..?
(2 x 1)
Rahul said:
6 years ago
Shouldn't it be 8C2 instead of 4C2? as two packs have 8 kings.
(1)
Mahi said:
9 years ago
Prince.
Formula for nCr = n!/(n-r)!/r!
Formula for nCr = n!/(n-r)!/r!
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