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

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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

[spoiler]OA=D[/spoiler]

Source: Economist GMAT Tutor

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by Brent@GMATPrepNow » Mon Apr 08, 2019 6:37 am
VJesus12 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
Let's use the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P( 2 different languages) = 1 - P(NOT 2 different languages)
So, we can write: P(2 different languages) = 1 - P(the 2 languages are the SAME)

P(the 2 languages are the SAME)
P(the 2 languages are the SAME) = P(both books are English OR both books are Spanish OR both books are Portuguese )
= P(both books are English) + P(both books are Spanish) + P(both books are Portuguese )
= P(1st book is English AND 2nd book is English) + P(1st book is Spanish AND 2nd book is Spanish) + P(1st book is Portuguese AND 2nd book is Portuguese)
= [P(1st book is English) x P(2nd book is English)] + [P(1st book is Spanish) x P(2nd book is Spanish)] + [P(1st book is Portuguese) x (P(2nd book is Portuguese)]
= [5/10 x 4/9] + [3/10 x 2/9] + [2/10 x 1/9]
= 20/90 + 6/90 + 2/90
= 28/90
= 14/45

So, P(2 different languages) = 1 - 14/45
= 31/45

Answer: D

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Brent
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by Scott@TargetTestPrep » Wed Apr 10, 2019 4:34 pm
VJesus12 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

[spoiler]OA=D[/spoiler]

Source: Economist GMAT Tutor
We can use the formula:

P(choosing two books of different language) = 1 - P(choosing two books of same language)

So our options are 2 English, 2 Spanish or 2 Portuguese books.

2 English books can be chosen in:

5C2 = (5 x 4)/2! = 10 ways

2 Spanish books can be chose in:

3C2 = 3 ways

2 Portuguese books can be chosen in:

2C2 = 1 way

The total ways to select 2 books from 10 is:

10C2 = (10 x 9)/2 = 45 ways

So P(choosing two books of same language) = (3 + 10 + 1)/45 = 14/45.

Thus, P(choosing two books of different language) = 1 - 14/45 = 31/45.

Alternate Solution:

If the first book is an English book (for which there is a 5/10 = 1/2 probability), the second book can be any of the 3 + 2 = 5 books among the 9 remaining books. Therefore, the probability of choosing an English book followed by a book in a different language is 1/2 x 5/9 = 5/18.

If the first book is a Spanish book (for which there is a 3/10 probability), the second book can be any of the 5 + 2 = 7 books among the 9 remaining books. Therefore, the probability of choosing an Spanish book followed by a book in a different language is 3/10 x 7/9 = 21/90 = 7/30.

If the first book is a Portuguese book (for which there is a 2/10 = 1/5 probability), the second book can be any of the 5 + 3 = 8 books among the 9 remaining books. Therefore, the probability of choosing an Portuguese book followed by a book in a different language is 1/5 x 8/9 = 8/45.

Therefore, the total probability of choosing two books in different languages is 5/18 + 7/30 + 8/45 = 25/90 + 21/90 + 16/90 = 62/90 = 31/45.

Answer: D

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