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

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:

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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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