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Difficulty:
35%
(medium)
Question Stats:
74% (01:59) correct
26%
(02:17)
wrong
based on 538
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On a shelf there are 6 hardback books and 2 paperback book. If we pick up 4 books at random, what is the probability that we pick up at least one paperback book?
A. 11/14
B. 5/7
C. 2/7
D. 3/14
E. 1/7
A. 11/14
B. 5/7
C. 2/7
D. 3/14
E. 1/7
On a shelf there are 6 hardback books and 2 paperback book. If we pick up 4 books at random, what is the probability that we pick up at least one paperback book?
A. 11/14
B. 5/7
C. 2/7
D. 3/14
E. 1/7
Let's find the probability of the opposite event and subtract this value from 1.
The opposite event would be if out of 4 books we pick all ll will be hardback: \(P(H=4)=\frac{C^4_6}{C^4_8}=\frac{15}{70}=\frac{3}{14}\).
Hence, \(P(P\geq{1})=1-\frac{3}{14}=\frac{11}{14}\).
Answer: A.
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A. 11/14
B. 5/7
C. 2/7
D. 3/14
E. 1/7
Let's find the probability of the opposite event and subtract this value from 1.
The opposite event would be if out of 4 books we pick all ll will be hardback: \(P(H=4)=\frac{C^4_6}{C^4_8}=\frac{15}{70}=\frac{3}{14}\).
Hence, \(P(P\geq{1})=1-\frac{3}{14}=\frac{11}{14}\).
Answer: A.
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EMPOWERgmatRichC
Expert
Major Poster
Joined: 19 Dec 2014
Last visit: 31 Dec 2023
Posts: 21,775
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Expert reply
Hi All,
The 'math' behind this question can be approached in a couple of different ways - and you can actually avoid using the Combination Formula altogether.
The prompt tells us that there are 6 hardcover books and 2 paperback books. We're asked for the probability of selecting AT LEAST one paperback book when we randomly select 4 books from the overall group of 8 books. Since the question focuses on getting AT LEAST one paperback book, we can determine what we DON'T want (meaning 0 paperback books) and subtract that result from 1 (to determine what we DO want).
Working one book at a time, the probability of NOT getting a paperback book is....
1st book = 6/8 chance of NOT getting a paperback
2nd book = 5/7 chance of NOT getting a paperback
3rd book = 4/6 chance of NOT getting a paperback
4th book = 3/5 chance of NOT getting a paperback
Thus, the probability of NOT getting a paperback book for the first 4 books is (6/8)(5/7)(4/6)(3/5). You should notice that the 6s and 5s 'cancel out', leaving us with...
(4)(3)/(8)(7) = 12/56 = 3/14
Thus, the probability of getting AT LEAST one paperback would be 1 - 3/14 = 11/14
Final Answer:
GMAT assassins aren't born, they're made,
Rich
The 'math' behind this question can be approached in a couple of different ways - and you can actually avoid using the Combination Formula altogether.
The prompt tells us that there are 6 hardcover books and 2 paperback books. We're asked for the probability of selecting AT LEAST one paperback book when we randomly select 4 books from the overall group of 8 books. Since the question focuses on getting AT LEAST one paperback book, we can determine what we DON'T want (meaning 0 paperback books) and subtract that result from 1 (to determine what we DO want).
Working one book at a time, the probability of NOT getting a paperback book is....
1st book = 6/8 chance of NOT getting a paperback
2nd book = 5/7 chance of NOT getting a paperback
3rd book = 4/6 chance of NOT getting a paperback
4th book = 3/5 chance of NOT getting a paperback
Thus, the probability of NOT getting a paperback book for the first 4 books is (6/8)(5/7)(4/6)(3/5). You should notice that the 6s and 5s 'cancel out', leaving us with...
(4)(3)/(8)(7) = 12/56 = 3/14
Thus, the probability of getting AT LEAST one paperback would be 1 - 3/14 = 11/14
Final Answer:
GMAT assassins aren't born, they're made,
Rich
General Discussion
ferrarih
OA:A
The probability of picking up at least one paperback book \(= 1 -\) The probability of picking up no paperback book \(= 1- \frac{C(6,4)}{C(8,4)}= 1-\frac{3}{14}=\frac{11}{14}\)
ScottTargetTestPrep
Expert
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 11 Sep 2026
Posts: 22,738
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert reply
Posts: 22,738
ferrarih
The phrase “at least one paperback” means “one or more paperbacks.” Thus, the only way we would NOT pick up at least one paperback would be if all 4 books were hardbacks. Thus, we can use the formula:
P(at least one paperback) = 1 - P(all hardbacks)
The number of ways to select all hardbacks is 6C4:
6! / (4! x 2!) = (6 x 5) / (2 x 1) = 15
The number of ways to select 4 books from 8 is 8C4:
8! / (4! x 4!) = (8 x 7 x 6 x 5) / (4 x 3 x 2 x 1) = 7 x 2 x 5 = 70
The probability of all hardbacks is 15/70 = 3/14 and thus P(at least on paperback) = 1 - 3/14 = 11/14.
Answer: A
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Archit3110
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Joined: 18 Aug 2017
Last visit: 12 Sep 2026
Posts: 8,816
Status:You learn more from failure than from success.
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1: 545 Q79 V79 DI73 
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GPA: 4
WE:Marketing (Energy)
GMAT Focus 2: 645 Q83 V82 DI81
Posts: 8,816
ferrarih
total possible ways = ( 1- ( 6/8* 5/7 * 4/6 * 3/5) ) = 11/14
IMO A
GMAT 1: 760 Q51 V42
Posts: 10,057
Hardback books = 6
Paperback books = 2
Total = 8
4 books picked at random: \(^8{C_4} =70 \)
At least one paperback: 1 - all 4 hardback
4 hardback books picked out of 6: \(^6{C_4} = ^6{C_2} = 15\)
Probability: \(\frac{15}{70} = \frac{3}{14}\)
At least one paperback: \(1 - \frac{3}{14} = \frac{11}{14}\)
Answer A
Paperback books = 2
Total = 8
4 books picked at random: \(^8{C_4} =70 \)
At least one paperback: 1 - all 4 hardback
4 hardback books picked out of 6: \(^6{C_4} = ^6{C_2} = 15\)
Probability: \(\frac{15}{70} = \frac{3}{14}\)
At least one paperback: \(1 - \frac{3}{14} = \frac{11}{14}\)
Answer A
On a shelf there are 6 hardback books and 2 paperback book. If we pick up 4 books at random, what is the probability that we pick up at least one paperback book?
A. 11/14
B. 5/7
C. 2/7
D. 3/14
E. 1/7
A. 11/14
B. 5/7
C. 2/7
D. 3/14
E. 1/7
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