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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 11 Apr 2021, 21:45
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35% (medium)

Question Stats:

74% (02:02) correct 26% (02:15) wrong based on 308 sessions

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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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D
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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 18 Apr 2021, 21:15
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Bunuel
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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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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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 11 Apr 2021, 22:36
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Bunuel
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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\(\frac{5_{C_1}*3_{C_1} + 3_{C_1}*2_{C_1} + 2_{C_1}*5_{C_1}}{10_{C_2}}\)
= \(\frac{31}{45}\)

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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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 11 Apr 2021, 23:28
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(5C1*3C1+3C1*2C1+5C1*2C1)/(10C2)= 31/45 OPTION D
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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 12 Apr 2021, 09:32
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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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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 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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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 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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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 16 May 2025, 23:30
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How would this change in case this were to be with replacement
unraveled
Bunuel
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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There are 10 books on a shelf: 5 English books, 3 Spanish books and 2 17 May 2025, 01:44
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SanchitSanjay
How would this change in case this were to be with replacement
unraveled
Bunuel
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.
Show more

If the selection is with replacement, the total logic changes slightly because the first book is replaced before picking the second. This means:

  • The total number of books remains 10 for both selections.
  • The second pick is independent of the first.

So:

  • Probability of selecting two English books: (5/10) * (5/10) = 25/100
  • Probability of selecting two Spanish books: (3/10) * (3/10) = 9/100
  • Probability of selecting two Portuguese books: (2/10) * (2/10) = 4/100

Total probability of selecting two books in the same language = 25/100 + 9/100 + 4/100 = 38/100

Therefore, the probability of selecting two books in different languages = 1 - 38/100 = 62/100 = 31/50.
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