Probability Questions and Answers – Cards – Set 2

This set of Aptitude Questions and Answers (MCQs) focuses on “Cards – Set 2”.

1. What is the probability of drawing two kings from a well shuffled pack of cards without replacement?
a) 5/663
b) 3/221
c) 1/221
d) 1/663
View Answer

Answer: c
Explanation:
Total cards = 52
Probability of 1st card being a king = 4/52 = 1/13
Probability of 2nd card being a king = 3/51 = 1/17
Total probability = 1/13 * 1/17
= 1/221

2. What is the probability of drawing two spades from a well shuffled pack of cards without replacement?
a) 4
b) 6
c) 8
d) 12
View Answer

Answer: a
Explanation:
Total cards = 52
Probability of 1st card being a spade = 13/52 = 1/4
Probability of 2nd card being a spade = 12/51 = 4/17
Total probability = 1/4 * 4/17
= 1/17

3. What is the probability of drawing one spade and one club from a well shuffled pack of cards without replacement?
a) 15/204
b) 13/204
c) 11/204
d) 9/204
View Answer

Answer: b
Explanation:
Total cards = 52
Probability of 1st card being a spade = 13/52 = 1/4
Probability of 2nd card being a club = 13/51
Total probability = 1/4 * 13/51
= 13/204
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4. What is the probability of drawing two numbered cards from a well shuffled pack of cards without replacement?
a) 105/221
b) 142/221
c) 90/221
d) 165/221
View Answer

Answer: a
Explanation:
Total cards = 52
Probability of 1st card being a number = 36/52 = 9/13
Probability of 2nd card being a number = 35/51
Total probability = 9/13 * 35/51
= 315/663 = 105/221

5. What is the probability of drawing two queens of diamond from a well shuffled pack of cards without replacement?
a) 1/4
b) 1
c) 0
d) 1/3
View Answer

Answer: c
Explanation:
Total cards = 52
Probability of 1st card being a diamond queen = 1/52
Probability of 2nd card being a diamond queen = 0/51 = 0
Total probability = 1/52 * 0
= 0

6. What is the probability of getting two ace cards from a well shuffled standard deck with replacement?
a) 8/169
b) 4/169
c) 16/169
d) 1/169
View Answer

Answer: d
Explanation:
Total cards = 52
Probability of 1st card being an ace = 4/52 = 1/13
Probability of 2nd card being an ace = 4/52 = 1/13
Total probability = 1/13 * 1/13
= 1/169

7. What is the probability of getting two heart cards from a well shuffled standard deck with replacement?
a) 1/16
b) 4/52
c) 8/52
d) 4/16
View Answer

Answer: a
Explanation:
Total cards = 52
Probability of 1st card being a heart = 13/52 = 1/4
Probability of 2nd card being a heart = 13/52 = 1/4
Total probability = 1/4 * 1/4
= 1/16
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8. What is the probability of getting two numbered cards from a well shuffled standard deck with replacement?
a) 13/169
b) 9/169
c) 81/169
d) 18/169
View Answer

Answer: c
Explanation:
Total cards = 52
Probability of 1st card being a number = 36/52 = 9/13
Probability of 2nd card being a number = 36/52 = 9/13
Total probability = 9/13 * 9/13
= 81/169

9. What is the probability of getting two non-numbered cards from a well shuffled standard deck with replacement?
a) 4/169
b) 16/169
c) 4/13
d) 13/169
View Answer

Answer: b
Explanation:
Total cards = 52
Probability of 1st card being a non-number = 16/52 = 4/13
Probability of 2nd card being a non-number = 16/52 = 4/13
Total probability = 4/13 * 4/13
= 16/169
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10. What is the probability of getting two black cards from a well shuffled standard deck with replacement?
a) 1/4
b) 1/8
c) 1/16
d) 1/13
View Answer

Answer: a
Explanation:
Total cards = 52
Probability of 1st card being a black card = 36/52 = 1/2
Probability of 2nd card being a black card = 36/52 = 1/2
Total probability = 1/2 * 1/2
= 1/4

To practice all aptitude questions, please visit “1000+ Quantitative Aptitude Questions”, “1000+ Logical Reasoning Questions”, and “Data Interpretation Questions”.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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