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John has on his shelf four books of poetry

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John has on his shelf four books of poetry [#permalink]

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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:
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Mike :-)
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Re: John has on his shelf four books of poetry [#permalink]

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New post 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]

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(4C2 * 2C2) / (10C2)

8/45. E
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John has on his shelf four books of poetry [#permalink]

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New post 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]

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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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Re: John has on his shelf four books of poetry [#permalink]

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New post 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]

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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.

Answer: E.

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

Answer: E.
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Re: John has on his shelf four books of poetry [#permalink]

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New post 05 Jul 2015, 02:57
Hi everyone here's the shortest way to tackle such questions - Probability approach
Total = 10 Books

4/10*4/9=8/45 (E) Consider to pick 9 books as total for the second choice, as there is 1 book less after the first pick
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Re: John has on his shelf four books of poetry [#permalink]

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New post 05 Jul 2015, 03:17
BrainLab wrote:
Hi everyone here's the shortest way to tackle such questions - Probability approach
Total = 10 Books

4/10*4/9=8/45 (E) Consider to pick 9 books as total for the second choice, as there is 1 book less after the first pick


hi, your answer is correct but approach is wrong...
how did you get 4 as numerator in both cases .. reference books are o two in number so it should be 2/9..
so to pick up two books randomly ans is 4/10*2/9, but these two books can be picked up in 2! ways within themselves..
so 4/10*2/9*2!=8/45..
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John has on his shelf four books of poetry [#permalink]

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New post 05 Jul 2015, 03:23
chetan2u wrote:
BrainLab wrote:
Hi everyone here's the shortest way to tackle such questions - Probability approach
Total = 10 Books

4/10*4/9=8/45 (E) Consider to pick 9 books as total for the second choice, as there is 1 book less after the first pick


hi, your answer is correct but approach is wrong...
how did you get 4 as numerator in both cases .. reference books are o two in number so it should be 2/9..
so to pick up two books randomly ans is 4/10*2/9, but these two books can be picked up in 2! ways within themselves..
so 4/10*2/9*2!=8/45..



The second 4 in the numerator is YOUR 2! ;) ... you can write it 2/9*2 or 4/9
I should have written it down in a more detailed way.. just too involved in such kind of questions in the last time.. that's why skipping some basic steps... ))
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Re: John has on his shelf four books of poetry [#permalink]

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New post 05 Jul 2015, 04:26
BrainLab wrote:
chetan2u wrote:
BrainLab wrote:
Hi everyone here's the shortest way to tackle such questions - Probability approach
Total = 10 Books

4/10*4/9=8/45 (E) Consider to pick 9 books as total for the second choice, as there is 1 book less after the first pick


hi, your answer is correct but approach is wrong...
how did you get 4 as numerator in both cases .. reference books are o two in number so it should be 2/9..
so to pick up two books randomly ans is 4/10*2/9, but these two books can be picked up in 2! ways within themselves..
so 4/10*2/9*2!=8/45..



The second 4 in the numerator is YOUR 2! ;) ... you can write it 2/9*2 or 4/9
I should have written it down in a more detailed way.. just too involved in such kind of questions in the last time.. that's why skipping some basic steps... ))


hi,
you did mention about denominator 10 and 9, so expected that you will not miss out mentioning about my 2! :wink: , after all it is more likely to be missed by students than 9 in denominator..
i know 2/9 * 2! is 4/9, but it is important to understand the reasoning behind it.. otherwise you could have straight way written 8/45 :wink: , after all it is equal to 4/10*4/9...
Anyway good shortest way to answer such Qs :) :)
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John has on his shelf four books of poetry [#permalink]

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BrainLab wrote:
chetan2u wrote:
BrainLab wrote:
Hi everyone here's the shortest way to tackle such questions - Probability approach
Total = 10 Books

4/10*4/9=8/45 (E) Consider to pick 9 books as total for the second choice, as there is 1 book less after the first pick


hi, your answer is correct but approach is wrong...
how did you get 4 as numerator in both cases .. reference books are o two in number so it should be 2/9..
so to pick up two books randomly ans is 4/10*2/9, but these two books can be picked up in 2! ways within themselves..
so 4/10*2/9*2!=8/45..



The second 4 in the numerator is YOUR 2! ;) ... you can write it 2/9*2 or 4/9
I should have written it down in a more detailed way.. just too involved in such kind of questions in the last time.. that's why skipping some basic steps... ))


Sorry, but both of you sound off in your approach to this.
The second 2 in the numerator appears because you need to account for the order in which you end up picking.
i.e.
p(novel) x p(ref) + p(ref) x p(novel)
(4/10 x 2/9) + (2/9 x 4/10)
or (4/10 x 2/9) x 2!
This is not the same as 4/10 x (2/9 x 2!)
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John has on his shelf four books of poetry [#permalink]

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New post 17 Dec 2017, 08:46
nyoebic wrote:
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:
http://gmatclub.com/forum/a-division-of ... 45433.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!


I'm still not sure I understand the logic here. We multiply by 2, which increases the probability that the event would occur. Why would the probability be higher (double) when we pick one book from two different genres, rather than picking two books from one genre? At the end of the day, we are still picking two books. In other words, why would the probability of picking one of one type of book and one of another type of book be higher than the probability of picking two of one type of book? As Usre123 mentioned, we can still pick those two books of the same genre in two different orders. Please help me understand the logic here - thanks!
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Re: John has on his shelf four books of poetry [#permalink]

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New post 18 Dec 2017, 13:30
mbah191 wrote:
I'm still not sure I understand the logic here. We multiply by 2, which increases the probability that the event would occur. Why would the probability be higher (double) when we pick one book from two different genres, rather than picking two books from one genre? At the end of the day, we are still picking two books. In other words, why would the probability of picking one of one type of book and one of another type of book be higher than the probability of picking two of one type of book? As Usre123 mentioned, we can still pick those two books of the same genre in two different orders. Please help me understand the logic here - thanks!

Dear mbah191,

This is Mike McGarry, author of the question. I'm happy to respond. :-)

The curious thing about this question is that, for the purpose of the question, all we know is that we have "four books of poetry, four novels, and two reference works." In other words, for the purpose of the question, we are considering the four books of poetry identical, the four novels identical, and the two reference works identical. This may or may not be the case with the real books, but this is how the problem is set up.

Let's pretend that all these books in each of the three categories are identical. Suppose, for some reason, John has four identical copies of the same poetry book, four identical copies of the same novel, and two identical copies of the same dictionary. The only way to arrive at the result of "two poetry books" would be to pick a poetry book on the first choice and another poetry book on the second choice. There's no other way to reach that result.

By contrast, to get the result "one novel and one reference book," it's possible to get to that result in two different ways:
(1) pick a novel first, then pick a reference work on the second choice
(2) pick a reference work first, then pick a novel on the second choice
Unlike the two-of-the-same case, there are two different routes that lead to this same result. If we figure out the probability of following one of these routes, we would have to double that probability to account for all the ways to arrive at that result.

Does all this make sense?
Mike :-)
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Re: John has on his shelf four books of poetry   [#permalink] 18 Dec 2017, 13:30
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