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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
30 Jun 2012, 07:35
Kudos
Bookmarks
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.
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.
22 Apr 2017, 16:18
Kudos
Bookmarks
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
