# Aptitude - Probability - Discussion

### Discussion :: 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?

[A].
 1 15
[B].
 25 57
[C].
 35 256
[D].
 1 221

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

 Shyam said: (Jan 19, 2011) How 2 kings out of 4 .?

 Rahul said: (May 31, 2011) 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: (Jul 19, 2011) @ Shyam Generally there are only four kings in a pack of 52 cards.

 Prince said: (Jul 24, 2011) (52 x 51) ---------..? how is this equation possible..? (2 x 1)

 S.Mounica said: (Jul 26, 2011) Its formula, nC2 = n*n-1/1*2 So we get it.

 Arun Sharma said: (Aug 5, 2011) 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: (Nov 8, 2011) @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: (Jun 12, 2012) 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: (Feb 16, 2013) 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: (May 29, 2013) 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: (Jun 4, 2013) @Santhu. 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: (Feb 12, 2014) 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: (May 13, 2014) @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.

 Dhruvil said: (Aug 19, 2014) 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).

 Bex said: (Feb 8, 2016) 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?

 Naveenaa said: (Oct 7, 2016) Could anyone kindly clarify at what type of situations we are about to use this nCn?

 Mahi said: (Dec 4, 2016) Prince. Formula for nCr = n!/(n-r)!/r!

 Tiger said: (Feb 14, 2017) 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.

 Vinayak said: (Jun 17, 2017) There are 12 kings in 52 cards. Then how to use 4 out 2. There is 12 out of 2.

 Ernie said: (Jun 22, 2017) @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).

 Srini said: (Jul 16, 2017) The probability of 2 out of 4 is 1/2. And the sample space is 2 /52. Is this correct or wrong can anyone explain?

 Arjun Singh said: (Apr 6, 2018) Here question is to draw two cards together. It means drawing two cards in one attempt/selection. which reduces the number of samples to 51. Explanation: let's assume we have tied two cards together randomly and then the total samples reduce to 51. Now out these, we are making one selection(as its a single drawing). which is 51c1. Let's assume the two selected cards to be kings and lets name cards as k1,k2,k3,k4, then total possible outcomes will be (k1,k2) (k1,k3) (k1,k4) (k2,k1) (k2,k3) (k2,k4) (k3,k1) (k3,k2) (k3,k4) (k4,k1) (k4,k2) (k4,k3) n(E) = 12. n(S) = 51. then P(E) = n(E)/ n(S) = 12/51 = 4/17. Am I right?

 Manupriya said: (Oct 19, 2018) There are 52 cards. In 52 cards 13 spade, 13 heart, 13 club, 13 Diamond is there in each of these 13 one will be king so in question we have to select 2 out of 4 kings. Therefore, the answer is 4c2/52c2.

 Gangadhar said: (Dec 10, 2018) There are 4 Kings so; (4/52) When you are taking 2nd card the total number of cards you have are 51 and out of 51 you have 3kings do (3/51). Finally, (4/52) * (3/51) = (1/221).

 Shreya said: (Jun 27, 2019) Here the question is the probability of getting kings only that is getting the same combination. Can we use (4/52) + (3/51)? Please tell me.

 Nikolay said: (Aug 5, 2019) @Shreya. It's (4/52) x (3/51) =1/221.

 Rahul said: (Aug 30, 2019) Shouldn't it be 8C2 instead of 4C2? as two packs have 8 kings.

 Gourang said: (May 24, 2020) @Rahul. When you choose a card from 52 you have 4 options so 4/52, now again you have to pick a card and you can't pick that same card you have picked before so 3 cards are left out of 51. So 3/51. Therefore 4/52 *3/51.