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When choose 3 out of 8 books, the probability of the most ex

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When choose 3 out of 8 books, the probability of the most ex  [#permalink]

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New post 10 Nov 2007, 19:39
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Please help me to understand this... I am always terrified when i face PROBABILITY questions...

Q. When choose 3 out of 8 books, the probability of the most expensive two books must be included?

Please helpe me kindly and thank you in advance...
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Re: (probability) 3 out of 8, the most expensive two include  [#permalink]

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New post 10 Nov 2007, 20:04
Vemuri wrote:
pretttyune wrote:
Please help me to understand this... I am always terrified when i face PROBABILITY questions...

Q. When choose 3 out of 8 books, the probability of the most expensive two books must be included?

Please helpe me kindly and thank you in advance...



Is the question complete?


yes... the question is .. In choosing 3 out of 8 books, the probability of the most expensive two books must be included?
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New post 08 Jan 2008, 21:58
walker wrote:
P=6C1/8C3=6*3*2/(8*7*6)=3/28


it is: consider the most expensive 2 books....than we have 6 slots free to combine the books. the total number of combs is 8C3 because we have a total of 8 books to combine in groups of three...prob=6/8*7=3/28
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Re: (probability) 3 out of 8, the most expensive two included...  [#permalink]

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New post 09 Jan 2008, 05:44
i thought about it like this.

probability = desired outcomes / total out comes

where desired outcomes is the number of outcomes of picking 3 out of 8 books, but the two most expensive have to be in there, so ive got (1C1)*(1C1)*(6C1) / 8C3 = 6/56 = 3/28
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Re: (probability) 3 out of 8, the most expensive two included...  [#permalink]

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New post 09 Jan 2008, 09:01
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i did it slightly differently.

(2/8 *1/7 * 6/6) * 3c1 = 3/28
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Re: (probability) 3 out of 8, the most expensive two included...  [#permalink]

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New post 25 Jan 2008, 04:19
walker wrote:
walker

Walker pls elucidate why not 2!6c1/8c3
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Re: (probability) 3 out of 8, the most expensive two included...  [#permalink]

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New post 25 Jan 2008, 04:35
AlexBon wrote:
walker wrote:
walker

Walker pls elucidate why not 2!6c1/8c3


\(p=\frac{C^2_2*C^6_1}{C^8_3}=\frac{1*6*3*2}{8*7*6}=\frac{3}{28}\)

or

\(p=\frac{C^2_2*C^6_1*P^3_3}{P^8_3}=\frac{1*6*3*2}{8*7*6}=\frac{3}{28}\)

or

\(p=\frac28*(\frac17*\frac66+\frac67*\frac16)+\frac68*\frac27*\frac16=\frac{3*12}{8*7*6}=\frac{3}{28}\)

I think 2! is permutation: \(P^2_2\). But you use \(C^8_3\) for all variants that mean ABC and BAC are the same variant. If we distinguish between ABC and BAC, we use \(P^8_3\) and \(P^3_3\)

Hope this help.
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Re: (probability) 3 out of 8, the most expensive two included...  [#permalink]

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New post 28 Sep 2009, 09:50
Q. When choose 3 out of 8 books, the probability of the most expensive two books must be included?


Soln: 6C1/8C3
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Re: (probability) 3 out of 8, the most expensive two included...  [#permalink]

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New post 17 Feb 2010, 02:44
pretttyune wrote:
Please help me to understand this... I am always terrified when i face PROBABILITY questions...

Q. When choose 3 out of 8 books, the probability of the most expensive two books must be included?

Please helpe me kindly and thank you in advance...


Out of 8 .. 2 books (which are most exp) should be considered in your selection.

No of ways the 2 Exp books can be selected = 2c2 = 1

remaining we need to choose one more book from the 6 left over books = 6c1 = 6

Total ways of selecting 3 books out of 8 = 8c3

Prob = 1X6 / 8c3 = 3/28
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Re: (probability) 3 out of 8, the most expensive two included...  [#permalink]

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New post 05 Aug 2013, 20:14
Probability of selecting most expensive out of 8 =1/8
Probability of selecting next most expensive out of 7 =1/7
And out of 6 left out can select any one out of 6 = 6c1/6c1
now probability of selecting 2 most expensive = 1/8*1/7*1*3!
=3/8(3! is because of arrangements)
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Re: When choose 3 out of 8 books, the probability of the most ex  [#permalink]

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Re: When choose 3 out of 8 books, the probability of the most ex   [#permalink] 15 Jan 2019, 01:44

When choose 3 out of 8 books, the probability of the most ex

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