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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2

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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2  [#permalink]

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New post 13 Jul 2015, 12:15
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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 Portuguese books. What is the probability of choosing 2 books in different languages?

A. 31/90
B. 3/10
C. 1/3
D. 31/45
E. 28/90

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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2  [#permalink]

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New post Updated on: 13 Jul 2015, 22:56
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1
reto wrote:
There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 Portuguese books. What is the probability of choosing 2 books in different languages?

A. 31/90
B. 3/10
C. 1/3
D. 31/45
E. 28/90


Probability = Favourable outcome / Total Outcome = 1- (Unfavourable outcome / Total Outcome)


METHOD-1

Favourable Outcomes

Case-1: One book is English (out of 5 English books) and one book is Spanish (out of 3 Spanish books) i.e. 5C1*3C1
Case-2: One book is Spanish (out of 3 Spanish books) and one book is Portuguese (out of 2 Portuguese books) i.e. 3C1*2C1
Case-3: One book is English (out of 5 English books) and one book is Portuguese (out of 2 Portuguese books) i.e. 5C1*2C1

Total favourable Outcomes = 5C1*3C1+3C1*2C1+5C1*2C1 = 15+6+10 = 31
Total Possible Outcomes = 10C2 = 45
i.e. Probability = 31/45

METHOD-2(Taking arrangement of books into account

Case-1: One book is English (out of 5 English books) and one book is Spanish (out of 3 Spanish books) i.e. 5*3*2! (with arrangement)
Case-2: One book is Spanish (out of 3 Spanish books) and one book is Portuguese (out of 2 Portuguese books) i.e. 3*2*2! (with arrangement)
Case-3: One book is English (out of 5 English books) and one book is Portuguese (out of 2 Portuguese books) i.e. 5*2*2! (with arrangement)

Total favourable Outcomes = 5*3*2!+3*2*2!+5*2*2! = 30+12+20 = 62
Total Possible Outcomes = 10*9 = 90
i.e. Probability = 62/90 = 31/4


METHOD-3

Unfavourable Outcomes

Case-1: Both selected books are English books (out of 5 English books) = 5C2
Case-2: Both selected books are Spanish books (out of 3 English books) = 3C2
Case-3: Both selected books are Portuguese books (out of 2 Portuguese books) = 2C2

Total Unfavourable Outcomes = 5C2+3C2+2C2 = 10+3+1 = 14
Total Possible Outcomes = 10C2 = 45
i.e. Probability = 1- (14/45) = 31/45
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Originally posted by GMATinsight on 13 Jul 2015, 22:48.
Last edited by GMATinsight on 13 Jul 2015, 22:56, edited 1 time in total.
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Re: There are 10 books on a shelf: 5 English books, 3 Spanish books and 2  [#permalink]

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New post 13 Jul 2015, 12:21
2
1
reto wrote:
There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 Portuguese books. What is the probability of choosing 2 books in different languages?

A. 31/90
B. 3/10
C. 1/3
D. 31/45
E. 28/90


ES + EP + SP = 2*5/10*3/9 + 2*5/10*2/9 + 2*3/10*2/9 = 31/45. Each time we are multiplying by 2 because each case has two ways of occurring. For example, English-Spanish can be picked as English first then Spanish as well as Spanish first then English.

Answer: D.
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Re: There are 10 books on a shelf: 5 English books, 3 Spanish books and 2  [#permalink]

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New post 13 Jul 2015, 12:43
Bunuel wrote:
reto wrote:
There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 Portuguese books. What is the probability of choosing 2 books in different languages?

A. 31/90
B. 3/10
C. 1/3
D. 31/45
E. 28/90


ES + EP + SP = 2*5/10*3/9 + 2*5/10*2/9 + 2*3/10*2/9 = 31/45. Each time we are multiplying by 2 because each case has two ways of occurring. For example, English-Spanish can be picked as English first then Spanish as well as Spanish first then English.

Answer: D.


Bunuel,
You multiply by 2 each case has two ways of occurring. But Is important to pick English first then Spanish or Spanish first then English? I think both are the same so we do not need to multiply by 2.

I hope you can help.

Thanks
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Re: There are 10 books on a shelf: 5 English books, 3 Spanish books and 2  [#permalink]

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New post 13 Jul 2015, 12:47
Mo2men wrote:
Bunuel wrote:
reto wrote:
There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 Portuguese books. What is the probability of choosing 2 books in different languages?

A. 31/90
B. 3/10
C. 1/3
D. 31/45
E. 28/90


ES + EP + SP = 2*5/10*3/9 + 2*5/10*2/9 + 2*3/10*2/9 = 31/45. Each time we are multiplying by 2 because each case has two ways of occurring. For example, English-Spanish can be picked as English first then Spanish as well as Spanish first then English.

Answer: D.


Bunuel,
You multiply by 2 each case has two ways of occurring. But Is important to pick English first then Spanish or Spanish first then English? I think both are the same so we do not need to multiply by 2.

I hope you can help.

Thanks


The point is that English-Spanish CAN occur in two ways, so we should account for both cases. It's not a matter of importance.
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Re: There are 10 books on a shelf: 5 English books, 3 Spanish books and 2  [#permalink]

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New post 13 Jul 2015, 13:00
The point is that English-Spanish CAN occur in two ways, so we should account for both cases. It's not a matter of importance.[/quote]

I used "important" by mistake. I mean is it a matter of arrangement? I still did not get your point. Can you please elaborate your point with an example ?

Thanks in advance for your support
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Re: There are 10 books on a shelf: 5 English books, 3 Spanish books and 2  [#permalink]

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New post 13 Jul 2015, 21:31
1
reto wrote:
There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 Portuguese books. What is the probability of choosing 2 books in different languages?

A. 31/90
B. 3/10
C. 1/3
D. 31/45
E. 28/90


My attempt -

Ways of choosing two books from 3 groups(one book from English AND other from Spanish OR one book from Spanish AND other from Portuguese OR one book from English AND other from Portuguese ) = 5C1*3C1+3C1*2C1+5C1*2C1 = 15+6+10 = 31

Total ways of choosing from 10 books = 10C2 = 45

Probability of choosing 2 books in different languages = 31/45.
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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2  [#permalink]

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New post 13 Jul 2015, 22:53
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Mo2men wrote:
The point is that English-Spanish CAN occur in two ways, so we should account for both cases. It's not a matter of importance.

I used "important" by mistake. I mean is it a matter of arrangement? I still did not get your point. Can you please elaborate your point with an example ?

Thanks in advance for your support



You are right about cause of your mistake

10C2 is 45 which is PURE SELECTION (WITHOUT ARRANGEMENT)

so We Don't need to take the arrangement of occurrence of two book English-Spanish for the given total outcomes (10C2) because total outcomes doesn't include arrangement

If you take two cases of spanish-English and English Spanish then you will have to take the total outcome in 10*9 = 90 ways
i.e. Total favourable Outcomes = 5*3*2!+3*2*2!+5*2*2! = 30+12+20 = 62
Total Possible Outcomes = 10*9 = 90
i.e. Probability = 62/90 = 31/45


I hope it helps!
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Re: There are 10 books on a shelf: 5 English books, 3 Spanish books and 2  [#permalink]

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New post 26 Jul 2015, 02:39
balamoon:

Why didnt we multiply by 2 for each selection of book??
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Re: There are 10 books on a shelf: 5 English books, 3 Spanish books and 2  [#permalink]

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New post 26 Jul 2015, 05:39
Shree9975 wrote:
balamoon:

Why didnt we multiply by 2 for each selection of book??


Why should we? Please explain. Keep in mind: order does not matter.
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Re: There are 10 books on a shelf: 5 English books, 3 Spanish books and 2  [#permalink]

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New post 26 Jul 2015, 05:57
Shree9975 wrote:
balamoon:

Why didnt we multiply by 2 for each selection of book??


To add to what reto has mentioned above, selecting an English book and then a Spanish book will NOT be different from selecting a Spanish book and then an English book.

You only need to select books from 2 books .

Consider the question this way:

The desired probability = 1 - (probability of selecting 2 books of the same language) = \(1- \frac{5C2+3C2+2C2}{10C2}\) = 31/45
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Re: There are 10 books on a shelf: 5 English books, 3 Spanish books and 2  [#permalink]

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New post 26 Jul 2015, 07:10
Bunuel wrote:
reto wrote:
There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 Portuguese books. What is the probability of choosing 2 books in different languages?

A. 31/90
B. 3/10
C. 1/3
D. 31/45
E. 28/90


ES + EP + SP = 2*5/10*3/9 + 2*5/10*2/9 + 2*3/10*2/9 = 31/45. Each time we are multiplying by 2 because each case has two ways of occurring. For example, English-Spanish can be picked as English first then Spanish as well as Spanish first then English.

Answer: D.


Hi,
isnt this Q slightly incomplete..
the wordings mean the prob of choosing two books of different languages..
I would take it to mean..
the 2 books of different languages can be chosen in following ways..
1) 3 books- one of one language and two of second language..
2) 4 books -2 of one language and two of second language..
3) 2 books- one of two languages each..
and so on

I think if the Q is supposed to mean what it actually wants to, the Q should be..
" what is the prob of the 2 books chosen to be of different languages?"
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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2  [#permalink]

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New post 03 Aug 2015, 03:05
The probaility of taking two same books:

(5/10*4/9)+(3/10*2/9)+(2/10*1/9)=28/90

1-28/90=62/90=31/45


Combination approach:

(5C1*3C1)+(5C1*2C1)+(3C1*2C1)/10C2 = 15+10+6/45=31/45

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Re: There are 10 books on a shelf: 5 English books, 3 Spanish books and 2  [#permalink]

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New post 03 Aug 2015, 05:40
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Required probability = (5C1 * 3C1) + (5C1 * 2C1) + (3C1 * 2C1)/10C2 = 15 + 10 + 6/45 = 31/45. Ans (D).
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Re: There are 10 books on a shelf: 5 English books, 3 Spanish books and 2  [#permalink]

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