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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

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

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09 Jan 2013, 15:42

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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 \(P2 = \frac{2}{10} * \frac{4}{9} = \frac{8}{90}\)

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.

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!

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".

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.

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

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05 Nov 2014, 11:23

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

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

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.. _________________

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

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05 Jul 2015, 04:23

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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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! , 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 , after all it is equal to 4/10*4/9... Anyway good shortest way to answer such Qs _________________

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

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19 Aug 2016, 14:17

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!)

gmatclubot

John has on his shelf four books of poetry
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19 Aug 2016, 14:17

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