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 1 of 4.
Shyam said:
1 decade ago
How 2 kings out of 4 .?
Rahul said:
1 decade ago
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:
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.
Prince said:
1 decade ago
(52 x 51)
---------..? how is this equation possible..?
(2 x 1)
---------..? how is this equation possible..?
(2 x 1)
S.Mounica said:
1 decade ago
Its formula, nC2 = n*n-1/1*2
So we get it.
So we get it.
Arun sharma said:
1 decade ago
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.
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:
1 decade ago
@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.
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:
1 decade ago
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?
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:
1 decade ago
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.
3/51 is the chance of getting 2nd King.
So, probability of getting 2 kings is:
4/52 * 3/51 = 1/221.
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.
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