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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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10 Nov 2007, 19:39
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Q. When choose 3 out of 8 books, the probability of the most expensive two books must be included?

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Re: (probability) 3 out of 8, the most expensive two include [#permalink]

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10 Nov 2007, 20:04
Vemuri wrote:
pretttyune wrote:

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

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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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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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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09 Jan 2008, 09:01
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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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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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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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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17 Feb 2010, 02:44
pretttyune wrote:

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

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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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)
Re: (probability) 3 out of 8, the most expensive two included...   [#permalink] 05 Aug 2013, 20:14
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