Bunuel wrote:
pretttyune wrote:
There are 8 books in a shelf that consist of 2 paperback books and 6 hardback books. If 4 books are selected at random without replacement, how many different combinations are there that at least one paperback book is selected?
A. 40
B. 45
C. 50
D. 55
E. 60
Thanks a lot for your help in advance..
It's almost always better to solve "at least one" combinations/probability questions with {at least one}={total}-{none}.
Total # of selections of 4 books out of 8: \(C^4_8=70\);
Total # of selections of 4 hardback books out of 6 (so none paperback): \(C^4_6=15\);
{at least one}={total}-{none}=70-15=55.
Answer: D.
Hi
Bunuel,
Using PB: Paperbacks ; HB : Hardbacks
If the question were -
"How many possible selections of 4 books include both the PB" then is why is my following solution wrong ?
Total possibility .: C(8,4)
Possibility of 2 PB = Total possibility - possibility of no PB => C(8,4)- C(6,4) [C(6,4) -> removing 2 PB from the lot will leave only 6 books. So selecting 4 books from 6].
So answer is C(8,4)- C(6,4) = 70-15 = 55