Aptitude - Probability - Discussion
Discussion Forum : Probability - General Questions (Q.No. 11)
11.
A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:
Answer: Option
Explanation:
Here, n(S) = 52.
Let E = event of getting a queen of club or a king of heart.
Then, n(E) = 2.
 P(E) = |
n(E) | = | 2 | = | 1 | . |
| n(S) | 52 | 26 |
Discussion:
55 comments Page 2 of 6.
Subrata Acharjee said:
1 decade ago
There are 13 cards in club and 13 also in heart.
so the probability of getting a queen of club is 1/13
and the probability of getting a king of heart is 1/13
so The probability of getting a queen of club or a king of heart is (1/13 + 1/13)
= 2/13
so the probability of getting a queen of club is 1/13
and the probability of getting a king of heart is 1/13
so The probability of getting a queen of club or a king of heart is (1/13 + 1/13)
= 2/13
Raj said:
1 decade ago
Not satisfied please help
since there are 13 cards of each club and heart category and each contain single king or queen then the answer should be 1/2.
ex:(13C1+13C1)/52C1
please help if wrong
since there are 13 cards of each club and heart category and each contain single king or queen then the answer should be 1/2.
ex:(13C1+13C1)/52C1
please help if wrong
KHAN said:
1 decade ago
See actually only one queen is there in club and only one king in heart therefore it will be : 1C1+1C1=2.
2/52=1/26.
2/52=1/26.
Madhu said:
1 decade ago
According to me, club And Heart contain 13, 13 cards. Given 1 queen in club or 1 king in heart. So the answer is n(s) =1.
And n(E) = 26. So p(E) =n(E) /n(s) = 1/26.
And n(E) = 26. So p(E) =n(E) /n(s) = 1/26.
Pooja said:
1 decade ago
Prob of 1 random selection n(S) = 52C1 = 52.
Queen of club = (13C1)/13 OR
King of heart = (13C1)/13.
So addition is n(E) = 2.
Finally,
p(E) = n(E)/n(S)
= 2/52 i.e 1/26.
Queen of club = (13C1)/13 OR
King of heart = (13C1)/13.
So addition is n(E) = 2.
Finally,
p(E) = n(E)/n(S)
= 2/52 i.e 1/26.
Rosy said:
1 decade ago
Hi, if asked as queen of club and king of heart, then can we do in this method
(13/26)*(12/25).
Here 1 queen of club selected so 13/26 and 1 king of heart selected so 12/25 Because already one is selected now they are the remaining.
Total cards 52c1.
So answer would be (13/26)*(12/25).
= 6/25.
So answer is (6/25)/52C1 = 312/25.
(13/26)*(12/25).
Here 1 queen of club selected so 13/26 and 1 king of heart selected so 12/25 Because already one is selected now they are the remaining.
Total cards 52c1.
So answer would be (13/26)*(12/25).
= 6/25.
So answer is (6/25)/52C1 = 312/25.
Jack said:
1 decade ago
Here's a simple way. I hope it helps:
Sample space is 52.
Now, in the entire pack of cards, only two cards satisfy the given condition of being either king of hearts of queen of clubs. Therefore, event space = 2
Probability = S/E = 2/52 = 1/26.
Sample space is 52.
Now, in the entire pack of cards, only two cards satisfy the given condition of being either king of hearts of queen of clubs. Therefore, event space = 2
Probability = S/E = 2/52 = 1/26.
Rekha said:
1 decade ago
In the question they specified that OR(+) so in the numerator we have 2 only.
Sajna said:
1 decade ago
In question, the term used is 'or', then how the probability of event became 2. (i.e, n (E) =2) ?
Shashank said:
1 decade ago
We have 1 queen of club and 1 queen of heart so,
[1C1(containing queen of club)+1c1(containing king of heart )]/52C1
[1+1]/52= 2/52=>1/26.
@Sajna. Simply "OR" means "addition" and "AND" means "MULTIPLICATION". Out of 52 cards we have only 1 queen of club and 1 king of heart. [1+1]/52.
[1C1(containing queen of club)+1c1(containing king of heart )]/52C1
[1+1]/52= 2/52=>1/26.
@Sajna. Simply "OR" means "addition" and "AND" means "MULTIPLICATION". Out of 52 cards we have only 1 queen of club and 1 king of heart. [1+1]/52.
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