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.
Ramakrishna Macharla said:
2 years ago
In a standard deck of 52 cards, there are 4 kings.
The probability of drawing the first king is 4/52. After drawing the first king, there are 3 kings left in a deck of 51 cards.
So, the probability of drawing the second king is 3/51.
To find the probability of both events happening together, you multiply the probabilities: (4/52) * (3/51) = 1/221.
Therefore, the probability of drawing both kings is 1/221.
The probability of drawing the first king is 4/52. After drawing the first king, there are 3 kings left in a deck of 51 cards.
So, the probability of drawing the second king is 3/51.
To find the probability of both events happening together, you multiply the probabilities: (4/52) * (3/51) = 1/221.
Therefore, the probability of drawing both kings is 1/221.
(22)
Pinky said:
2 years ago
Here they asked about the probability of 2 kings out of 52 cards.
No.of kings in a deck of cards = 4
P(E)= no of favourable outcomes/no.of possible outcomes.
So, 4/52 = 1/13.
No.of kings in a deck of cards = 4
P(E)= no of favourable outcomes/no.of possible outcomes.
So, 4/52 = 1/13.
(4)
Purva said:
3 years ago
Two cards are drawn together dependent event.
P(A and B) = P(A) * P(B/A).
= 4/52 * 3/51 = 1/221.
P(A and B) = P(A) * P(B/A).
= 4/52 * 3/51 = 1/221.
(7)
Alok said:
3 years ago
As drawing of cards are dependent events, I think 4c1*3c1/52c2 will be correct.
(1)
Gourang said:
5 years ago
@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.
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.
(1)
Rahul said:
6 years ago
Shouldn't it be 8C2 instead of 4C2? as two packs have 8 kings.
(1)
Nikolay said:
6 years ago
@Shreya.
It's (4/52) x (3/51) =1/221.
It's (4/52) x (3/51) =1/221.
(1)
Shreya said:
6 years ago
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.
Can we use (4/52) + (3/51)? Please tell me.
(3)
Gangadhar said:
7 years ago
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).
(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).
(6)
Manupriya said:
7 years ago
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.
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.
(2)
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