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15 Oct 2020, 10:30
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
46% (02:54) correct
54%
(03:25)
wrong
based on 89
sessions
History
Date
Time
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Not Attempted Yet
There are 4 fiction books and 6 non-fiction books in a reading list. 3 of the non-fiction are biographies. Each person is required to select 3 books from the list. What is the probability that at least one fiction book, and at most one biography are selected?
A. 1/5
B. 4/15
C. 1/3
D. 2/3
E. 11/15
A. 1/5
B. 4/15
C. 1/3
D. 2/3
E. 11/15
15 Oct 2020, 20:28
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Bunuel
Total ways - \(10C3=\frac{10*9*8}{(3!)}=120\)
So we have to pick at least 1 of 4 fiction books and 0 or 1 biographies.
a) Say all 3 are fiction - 4C3 or 4 ways.
b)Next only 2 are fiction - 4C2*(10-4)=6*6=36
c)Only 1 is fiction and no biography- 4C1*3C2=12
d)Only 1 is fiction and 1 is biography - 4C1*3C1*3C1=36
Total with restrictions = 4+36+12+36=88
Probability =\(\frac{88}{120}=\frac{11}{15}\)
E
15 Oct 2020, 21:23
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Bunuel
Explantion:
probability that at least one fiction book, and at most one biography are selected
=P(1F1B +2F1B+3F1B+4F1B)
=(4C1/10C1)*3/9 + (4C2/10C2)*3/8 + (4C3/10C3)*3/7 + (4C4/10C4)*3/6
=2/15 + 1/20 +1/70 +1/420
=84/420
=1/5
hence A is the correct answer IMO
