On a shelf there are 6 hardback books and 2 paperback book.

Question

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

Bunuel · Accepted Answer

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.

EMPOWERgmatRichC · Answer

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:  A

GMAT assassins aren't born, they're made,
Rich

Princ · Answer

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 · Answer

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

Archit3110 · Answer

total possible ways = ( 1- ( 6/8* 5/7 * 4/6 * 3/5) ) = 11/14
IMO A

MathRevolution · Answer

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

MBAHOUSE · Answer

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

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