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

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by varun289 » Sun Dec 16, 2012 5:02 am
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?
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by jkaustubh » Sun Dec 16, 2012 5:15 am
The answer is 31/45

The total number of ways of choosing 2 books out of 10 books is C(10,2) i.e 45

We have three cases:

Case I -
One E, One S

Total number of ways = C(5,1)*C(3,1)=5*3=15

Case II -
One E, One P

Total number of ways = C(5,1)*C(2,1)=5*2=10

Case III -
One P, One S

Total number of ways = C(3,1)*C(2,1)=3*2=6


Now the required probability = [15+10+6]/45 = 31/45
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by varun289 » Sun Dec 16, 2012 6:54 am
can we think like this -

3 color to choose ,
first choice - 1/3

second with other that first - 2/3

combine probability - 1/3*2/3=2/9

can any body reply me
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by Brent@GMATPrepNow » Sun Dec 16, 2012 9:37 am
varun289 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?
We can also use probability rules to solve this question.

P(2 different languages) = 1 - P(2 same languages)

P(2 same languages) = P(2 English OR 2 Spanish OR 2 Portuguese)
= P(2 English) + P(2 Spanish) + P(2 Portuguese)
= P(English 1st and English 2nd) + P(Spanish 1st and Spanish 2nd ) + P(Portuguese 1st and Portuguese 2nd)
= (5/10) x (4/9) + (3/10) x (2/9) + (2/10) x (1/9)
= 20/90 + 6/90 + 2/90
= 28/90

So, P(2 different languages) = 1 - P(2 same languages)
= 1 - 28/90
= 62/90
= [spoiler]31/45[/spoiler]

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Brent
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by Brent@GMATPrepNow » Sun Dec 16, 2012 9:41 am
varun289 wrote:can we think like this -

3 color to choose ,
first choice - 1/3

second with other that first - 2/3

combine probability - 1/3*2/3=2/9

can any body reply me
Hi Varun,

Your solution doesn't take into account that there are different numbers of different language books.
For example, the probability of the first selection being an English book is 1/2 while the probability of the first selection being a Spanish book is 3/10

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Brent
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by bnpetteway » Thu Jan 10, 2013 6:21 pm
@brent, why did you subtract 1 when it asked for two. I did everything the exact same way you did, except for the final step.
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by Brent@GMATPrepNow » Thu Jan 10, 2013 11:20 pm
We're looking for P(2 different languages).

To find this probability, we use the fact that P(2 different languages) = 1 - P(2 same languages)

I hope that helps.

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Brent
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by sonalibhangay » Fri Jan 11, 2013 8:08 am
We can also calculate by the Permutations & Combinations Method

Total ways of selecting 2 Books from 10 will be 10 C 2 = 45

Out of these, to get books in different languages, we need to subtract the ways where we would be selecting 2 books in same language
which will be following:

Ways to select books in different languages = Total ways - Ways to select books in same language

45 - (5C2 OR 3C2 OR 2C2) =

45- (10 + 3 + 1) = 31

Divide by the total number of ways to obtain probability of selecting books in different languages

[spoiler]31/45.[/spoiler]
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by Brent@GMATPrepNow » Sun Jan 20, 2013 10:18 am
sonalibhangay wrote:We can also calculate by the Permutations & Combinations Method

Total ways of selecting 2 Books from 10 will be 10 C 2 = 45

Out of these, to get books in different languages, we need to subtract the ways where we would be selecting 2 books in same language
which will be following:

Ways to select books in different languages = Total ways - Ways to select books in same language

45 - (5C2 OR 3C2 OR 2C2) =

45- (10 + 3 + 1) = 31

Divide by the total number of ways to obtain probability of selecting books in different languages

[spoiler]31/45.[/spoiler]
Perfect!

Cheers,
Brent
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