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GMATT73
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Math, Physics, and Chemistry books are stored on library 16 Nov 2005, 08:07
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Math, Physics, and Chemistry books are stored on library shelf that can accomodate 25 books. Currently, 20% of the shelf spots remain empty. There are twice as many math books as physics books and the number of physics books is 4 greater than that of the chemistry books. If 1 textbook selected at random is removed from the shelf to be shipped to another library, what is the probability of selecting a chemistry book among the remaining books?

A. 1/190
B. 1/19
C. 2/19
D. 9/95
E. 1/10

Figuring out the number of each book type on the shelf is no problem, but determining which "random" book was removed is the tricky part (for me).

Please show your work.



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Math, Physics, and Chemistry books are stored on library 16 Nov 2005, 08:30
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this is a good one.

i got E.

let x be number of physics book

so number of math books will be 2x and number of chemistry book will be x-4

so x+2x+x-4= 25*4/5
so x = 6.

so number of chemistry book is 2 and number of math book is 12

now the probability of the book shiped is chemistry is 2/20
and the probability of the book shiped is not chemistyr is 18/20

the probability of selecting a chemistry book among the remaining books
is 2/20*1/19 + 18/20*2/19 = 38/(20*19) = 1/10

so E is the answer.
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Math, Physics, and Chemistry books are stored on library 16 Nov 2005, 14:35
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E



20=2(c+4)+c+4+c
c=2,M=12,P6

probability of chemistery book =
If 1st book removed was chemistery then probability = 2/20*1/19=2/19*20

if 1st book removed was not chemistery = 2/19*18/19 = 36/19*20

total probability = case1 + case 2 = 38/19*20=2/20=1/10
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Math, Physics, and Chemistry books are stored on library 17 Nov 2005, 00:20
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why not C(2,1)/C(20,1)=2/20=1/10?
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GMATT73
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Math, Physics, and Chemistry books are stored on library 17 Nov 2005, 08:03
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nakib77
this is a good one.

i got E.

let x be number of physics book

so number of math books will be 2x and number of chemistry book will be x-4

so x+2x+x-4= 25*4/5
so x = 6.

so number of chemistry book is 2 and number of math book is 12

now the probability of the book shiped is chemistry is 2/20
and the probability of the book shiped is not chemistyr is 18/20

the probability of selecting a chemistry book among the remaining books
is 2/20*1/19 + 18/20*2/19 = 38/(20*19) = 1/10

so E is the answer.
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Nakib> where did you get that 2/19 figure from??
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nakib77
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Math, Physics, and Chemistry books are stored on library 17 Nov 2005, 08:15
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GMATT73


the probability of selecting a chemistry book among the remaining books
is 2/20*1/19 + 18/20*2/19 = 38/(20*19) = 1/10

so E is the answer.
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Nakib> where did you get that 2/19 figure from??[/quote]

if the shipped book is not chemistry, having a probability of 18/20, so there will be still 2 chemistry books left out of the 19. so the probability of selecting the chemistry will be 2/19.
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Math, Physics, and Chemistry books are stored on library 17 Nov 2005, 11:09
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oops... double posting.
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GMATT73
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Math, Physics, and Chemistry books are stored on library 17 Nov 2005, 22:56
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Got it! Indeed the OE is E as Nakib has so cleverly pointed out.
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ywilfred
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Math, Physics, and Chemistry books are stored on library 17 Nov 2005, 23:15
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m = 2p
p = 4+c
so m = 8+2c

# of books on the shelf = 20 books
so 8 + 2c + 4 + c + c = 20
4c = 8
c = 2

p = 6
m = 12

Probability = P(m and c) OR P(p and c) or P(c and c) = 12/20*2/19 + 6/20*2/19 + 2/20*1/19
= 1/10

Ans: E
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Math, Physics, and Chemistry books are stored on library 19 Nov 2005, 04:48
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This is a good one! I choose E - 1/10

We can calculate that Maths-12, Physics-6 and Chemistry - 2 books.

If the randomly selected book is a chemistry book then - 2/20 x 1/19 is the probablity of selecting chemistry from the remaining.

If the randomly selected book is non-chmeistry book then - 18/20 x 2/19 is the probablity of selecting chemistry from the remaining.

So the final probability is 1/10.
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Math, Physics, and Chemistry books are stored on library 19 Nov 2005, 05:54
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ywilfred
m = 2p
p = 4+c
so m = 8+2c

# of books on the shelf = 20 books
so 8 + 2c + 4 + c + c = 20
4c = 8
c = 2

p = 6
m = 12

Probability = P(m and c) OR P(p and c) or P(c and c) = 12/20*2/19 + 6/20*2/19 + 2/20*1/19
= 1/10

Ans: E
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Wilfred> I like your process, but the math calculates to 20/190=2/19 (not 1/10). What am I missing?



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