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# John has on his shelf four books of poetry

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Magoosh GMAT Instructor
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John has on his shelf four books of poetry [#permalink]  09 Jan 2013, 14:15
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John has on his shelf four books of poetry, four novels, and two reference works. Suppose from these ten books, we were to pick two books at random. What is the probability that we pick one novel and one reference work?
(A) 1/2
(B) 2/5
(C) 3/10
(D) 7/20
(E) 8/45

For a full discussion of probability and counting questions, as well as a complete solution to this question, see:
http://magoosh.com/gmat/2013/gmat-proba ... echniques/

Mike
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Mike McGarry
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Re: John has on his shelf four books of poetry [#permalink]  09 Jan 2013, 14:42
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You can choose 1 Novel book and 1 Reference work in two ways:
If you choose Novel first and then Reference -> Probability $$P1 = \frac{4}{10} * \frac{2}{9} = \frac{8}{90}$$
If you choose Reference first and then Novel -> Probability $$P2 = \frac{2}{10} * \frac{4}{9} = \frac{8}{90}$$

Probability(1 novel and 1 reference work) = $$P1 + P2 = \frac{8}{90}+\frac{8}{90}=\frac{8}{45}.$$

Hence choice (E)
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Re: John has on his shelf four books of poetry [#permalink]  10 Jan 2013, 07:52
(4C2 * 2C2) / (10C2)

8/45. E
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John has on his shelf four books of poetry [#permalink]  19 Sep 2014, 10:27
hello,
Why can't we do it this way: 4/10 *2/9= 4/45?
And if we are multiplying by 2 to change the order, then why didn't we do the same for this question:
a-division-of-a-company-consists-of-seven-men-and-five-women-145433.html

Here we used the same approach: 5/12 * 4/11 = 5/33. I'm very confused why the first question has been multiplied by two and not this one as well. After all, there are two ways of picking the women too.
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Re: John has on his shelf four books of poetry [#permalink]  12 Oct 2014, 10:18
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usre123 wrote:
hello,
Why can't we do it this way: 4/10 *2/9= 4/45?
And if we are multiplying by 2 to change the order, then why didn't we do the same for this question:
a-division-of-a-company-consists-of-seven-men-and-five-women-145433.html

Here we used the same approach: 5/12 * 4/11 = 5/33. I'm very confused why the first question has been multiplied by two and not this one as well. After all, there are two ways of picking the women too.

This is because this question asks us to pick two separate books. If the question asked us the probability to pick 2 novels then you wouldn't multiply with 2. Similarly, the question you are referring to asks you the probability of picking 2 women. Therefore you don't multiply by 2. Had it asked you the probability of picking one man and one woman then you would multiply by 2 because you could pick the man first or pick the man second. When you pick only women, it doesn't matter what's first or second because they both are women. Hope that helps!
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Probability PS question [#permalink]  04 Nov 2014, 06:09
Hello,
I can't remember where I picked up this question. I have tried to solve it, researching and studying about probability. However I could not figure out the solution.

The question is :
John has on his shelf four books of poetry, four novels, and two reference works.  Suppose from these ten books, we were to pick two books at random.  What is the probability that we pick one novel and one reference work?
1. 1/2
2. 2/5
3. 3/10
4. 7/20
5. 8/45

The solution I have tried is "4/10*2/9 / 10C2" where the numerator consists of "# of novels/total books * # of reference work / total books-1" and the denominator is "total # of possible combinations to select 2 books at random out of 10 books".
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Re: John has on his shelf four books of poetry [#permalink]  04 Nov 2014, 06:12
Expert's post
h31337u wrote:
Hello,
I can't remember where I picked up this question. I have tried to solve it, researching and studying about probability. However I could not figure out the solution.

The question is :
John has on his shelf four books of poetry, four novels, and two reference works.  Suppose from these ten books, we were to pick two books at random.  What is the probability that we pick one novel and one reference work?
1. 1/2
2. 2/5
3. 3/10
4. 7/20
5. 8/45

The solution I have tried is "4/10*2/9 / 10C2" where the numerator consists of "# of novels/total books * # of reference work / total books-1" and the denominator is "total # of possible combinations to select 2 books at random out of 10 books".

Merging topics.

Please refer to the discussion above and ask if anything remains unclear.

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Re: John has on his shelf four books of poetry [#permalink]  05 Nov 2014, 10:23
I agree with one of the above posters. I think the correct answer should 4/5. Once we pick a book, it's no longer on the shelf. So (4/10) * (2/9) = 4/45

incorrect?
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Re: John has on his shelf four books of poetry [#permalink]  06 Nov 2014, 05:14
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JackSparr0w wrote:
John has on his shelf four books of poetry, four novels, and two reference works. Suppose from these ten books, we were to pick two books at random. What is the probability that we pick one novel and one reference work?
(A) 1/2
(B) 2/5
(C) 3/10
(D) 7/20
(E) 8/45

I agree with one of the above posters. I think the correct answer should 4/5. Once we pick a book, it's no longer on the shelf. So (4/10) * (2/9) = 4/45

incorrect?

When we are picking two books, one novel and one reference work, we could either pick a novel first and then a reference book or pick a reference book and then a novel. Therefore the answer is 4/10*2/9 + 2/10*4/9 = 8/45.

Or, use combinations: $$P=\frac{C^1_4*C^1_2}{C^2_{10}}=\frac{8}{45}$$.

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Re: John has on his shelf four books of poetry   [#permalink] 06 Nov 2014, 05:14
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