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

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Bunuel
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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 [#permalink] New post   11 Apr 2021, 21:45
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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 Portuguese books. If selecting two books at random and without replacement, 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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Re: There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 [#permalink] New post   18 Apr 2021, 21:15
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Bunuel wrote:
There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 Portuguese books. If selecting two books at random and without replacement, 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] New post   11 Apr 2021, 22:36
Bunuel wrote:
There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 Portuguese books. If selecting two books at random and without replacement, 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

\(\frac{5_{C_1}*3_{C_1} + 3_{C_1}*2_{C_1} + 2_{C_1}*5_{C_1}}{10_{C_2}}\)
= \(\frac{31}{45}\)

OR
\(1 - \frac{5_{C_2} + 3_{C_2} + 2_{C_2}}{10_{C_2}}\)
= \(1- \frac{14}{45}\)
= \(\frac{31}{45}\)

Answer D.
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Re: There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 [#permalink] New post   11 Apr 2021, 23:28
(5C1*3C1+3C1*2C1+5C1*2C1)/(10C2)= 31/45 OPTION D
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Re: There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 [#permalink] New post   12 Apr 2021, 09:32
P(2 different books)= 1- P(Combinations of equal books/Total combinations)

P(Combinations of equal books/Total combinations) = C(5,2) (total combinations of 2 equal english books among 5 English books) + C(3,2) (total combinations of 2 equal spanish books among 3 spanish books ) + C(2,2) (total combinations of 2 equal portuguese books among 2 portuguese books ) =14

Total combinations C(10,2)=45

P(2 different books)= 1-14/45= 31/45 OPTION D
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Re: There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 [#permalink] New post   18 Apr 2021, 21:57
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Probability of 2 diff books = 1- prob of 2 same books

Probability of 2 same books= 5/10*4/9 + 3/10*2/9 +2/10*/9= 28/90

Probability of 2 diff books = 1- (28/90)= 62/90 = 31/45

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Re: There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 [#permalink] New post   18 Oct 2022, 10:50
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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 Portuguese books. If selecting two books at random and without replacement, 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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Re: I saw these on another forum-I think business week. 1. There [#permalink] New post   11 Jan 2024, 07:43
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Re: I saw these on another forum-I think business week. 1. There [#permalink]
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