Because in apack of cards 4 kings are there. So selecting 2 kings out of 4 is 4c2. Generally in apack of cards the 4 suits are aces, kings, queens, jacks.
Ajith Kumar said:
(Tue, Jul 19, 2011 12:05:23 PM)
@ Shyam
Generally there are only four kings in a pack of 52 cards.
Prince said:
(Sun, Jul 24, 2011 07:24:45 PM)
(52 x 51)
---------..? how is this equation possible..?
(2 x 1)
S.Mounica said:
(Tue, Jul 26, 2011 10:16:39 AM)
Its formula, nC2 = n*n-1/1*2
So we get it.
Arun Sharma said:
(Fri, Aug 5, 2011 11:34:59 PM)
I am really confused about this answer. I think the logic of 2 kings out of 4 is wrong. There are 13 types of cards in a pack and you are looking for the two cards that you are pulling out to be of one type.
So it should be (13x12)/(2x1) - this is the probability that the two cards you pull will be of one type. This will have to be divided by the sample space.
So it will be 156/2 = 78.
Hence 78/1376 = 39/692.
Swetha said:
(Tue, Nov 8, 2011 03:36:51 PM)
@Arun Sharma
There are 13 types of cards.,but the question asked is both are kings. We have only 4 kings. Therefore we must take 2 kings out of 4 kings.
Santhu said:
(Tue, Jun 12, 2012 10:34:25 PM)
Can anyone clarify my doubt?
In pack of card there are 4 kings once if we draw king then 4C1 then remaining kings are 3 2nd king we need to take from 3 so 3C1.
So answer is 4C1*3C1/52C2 is it correct?
Rohit said:
(Sat, Feb 16, 2013 08:19:14 PM)
4/52 is the chance of getting 1st King.
3/51 is the chance of getting 2nd King.
So, probability of getting 2 kings is:
4/52 * 3/51 = 1/221.
Lola said:
(Wed, May 29, 2013 12:43:41 PM)
The n(k) = 4, so the p(getting a king)=4c1, or 4/52 : p(getting 2 king) = 4c2. NB: combination mean selection.
Gaurav Garg said:
(Tue, Jun 4, 2013 12:05:58 PM)
@Santhu.
We are not selecting cards from only 4 kings, we are selecting them from all 52 cards And next time from 51 cards.
n(E) = 4C2 =