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
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.
Darshan said:
4 years ago
They have clearly mentioned Queen of club and king of hearts. It is common sense there is only one queen of club and king of heart.
@Abhishek
13C1 means your selecting all thirteen cards.
13C1 Gives 13,
1card cannot be selected in 13 ways.
@Abhishek
13C1 means your selecting all thirteen cards.
13C1 Gives 13,
1card cannot be selected in 13 ways.
(7)
Abhishek Kumar said:
5 years ago
In this question, there are 13 cards in the club and we have to draw the queen of the club then 13C1 and same with the king so for that also 13C1 and because of OR operation (13C1+13C1).
So, probability will be (13C1+13C1)/52C1 = 1/2.
So, probability will be (13C1+13C1)/52C1 = 1/2.
(16)
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
Deepti said:
1 decade ago
There are 1 queen in club n 1 king in heart, in question we have 1 queen from club or (+) 1king from heart which implies 1 + 1 = 2, we divide it by 52 i.e. total cards and get an answer 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.
Arijit Biswas said:
1 decade ago
I believe OR means addition and AND means multiplication. Hence we have to add the two events i.e 1+1=2 and divide by sample space which is equal to 52 giving answer 1/26.
Unnati Gupta said:
8 years ago
Ans. is : 1/13.
n(s) = 52C1 = 52; Because we need to find the probability of getting a queen of club OR-( / ) a king of spade.
n(E) = 4C1 = 4.
n(P) = 4/52 = 1/13.
n(s) = 52C1 = 52; Because we need to find the probability of getting a queen of club OR-( / ) a king of spade.
n(E) = 4C1 = 4.
n(P) = 4/52 = 1/13.
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.
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