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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